Magnetic Field Exercise HC Verma Solutions Ch-34 Vol-2
Magnetic Field Exercise HC Verma Solutions Ch-34 Vol-2 Concept of Physics Vol-2 for ISC Class-12. Step by Step Solution of Exercise Questions of Ch-34 Magnetic Field HC Verma Concept of Physics. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Magnetic Field Exercise HC Verma Solutions Ch-34 Vol-2
| Board | ISC and other board |
| Publications | Bharti Bhawan Publishers |
| Ch-34 | Magnetic Field |
| Class | 12 |
| Vol | 2nd |
| writer | H C Verma |
| Book Name | Concept of Physics |
| Topics | Solution of Exercise Questions |
| Page-Number | 230, 231, 232, 233, 234, 235 |
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Magnetic Field Exercise HC Verma Questions Solutions
(HC Verma Ch-34 Concept of Physics Vol-2 for ISC Class-12)
(Page-230)
Question-1 :-
An alpha particle is projected vertically upward with a speed of 3.0 × 104 km s−1 in a region where a magnetic field of magnitude 1.0 T exists in the direction south to north. Find the magnetic force that acts on the α-particle.
Answer-1 :-
Given:
Upward speed of the alpha particle,
v = 3 × 104 km/s
= 3 × 107 m/s
Magnetic field,
B = 1.0 T
The direction of the magnetic field is from south to north.
Charge of the alpha particle,
q = 2e,
where e is the charge of an electron.
q = 2 × 1.6 × 10−19 C,
Magnetic force acting on the α-particle,
= 9.6 × 10−19 N , towards west
The direction of magnetic force can be found using Fleming’s left-hand rule.
Question-2 :-
An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic field of strength 1.0 × 10−7 T exists in the vertically upward direction.
(a) Will the electron deflect towards the right or left of its motion?
(b) Calculate the sideways deflection of the electron while travelling through 1 m. Make appropriate approximations.
Answer-2 :-
Given:
The kinetic energy of the electron projected in the horizontal direction, K.E = 10 keV = 1.6 × 10−15 J
Magnetic field,
B = 1 × 10−7 T
The direction of magnetic field is vertically upward.
(a) The direction can be found by the right-hand screw rule. So, the electron will be deflected towards left.
(b) Kinetic energy,
t= Time taken to cross the magnetic field
As there is no force acting on the electron in the horizontal direction, the velocity of the electron remain constant in this direction.
So, the time taken to cross a distance of 1m in the horizontal direction in the magnetic field,
s = 0.0148 = 1.5 × 10−2 cm
Question-3 :-
A magnetic field of (4.0×10‾3k‾) T exerts a force of (4.0i‾+3.0j‾)×10‾10N on a particle with a charge of 1.0 × 10−9 C and going in the x − y plane. Find the velocity of the particle.
Answer-3 :-
Magnetic field, B = (4×10 x −3K^)T
Force exerted by the magnetic field on the charged particle, F = (4i^+3j^)× 10−10 N
Charge of the particle, q = 1 × 10−9 C
As per the question, the charge is going in the X-Y plane.
So, the x-component of force, Fx = 4 × 10−10 N
and the y-component of force, Fy = 3 × 10−10 N
Considering the motion along x-axis:-
Fx = qvy × B
On putting the respective values,
we get:-
vy = 100 m/s
Motion along y-axis:-
Fy = qvx × B
⇒ vx = 75 m/s
Thus,
total velocity =(-75i‾ + 100j‾)m/s
Question-4 :-
An experimenter’s diary reads as follows: “A charged particle is projected in a magnetic field of (7.0i‾ – 3.0j‾) × 10−3 T. The acceleration of the particle is found to be (xi‾ + 7.0j‾) The number to the left of i in the last expression was not readable. What can this number be?
Answer-4 :-
Magnetic field,
B = (7.0i − 3.0j) × 10−3 T
Acceleration of the particle,
a = (xi + 7j) × 10−6 m/s2
We have denoted the unidentified number as x.
B and a are perpendicular to each other. (Because magnetic force always acts perpendicular to the motion of the particle)
So, the dot product of the two quantities should be zero.
That is,
B.a = 0
⇒ 7x × 10−3 × 10−6 − 3 × 10−3 × 7 × 10−6 = 0
⇒ 7x − 21 = 0
x=21/7
=3
Hence,
Acceleration of the particle is (3i + 7j) × 10−6 m/s2.
Question-5 :-
A 10 g bullet with a charge of 4.00 μC is fired at a speed of 270 m s−1 in a horizontal direction. A vertical magnetic field of 500 µT exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations.
Answer-5 :-
Mass of the bullet,
m = 10g
Charge of the bullet,
q = 4.00 μC
Speed of the bullet in horizontal direction,
v = 270 m/s
Vertical magnetic field,
B = 500 μT
Distance travelled by the bullet,
d = 100 m
Magnetic force,
F‾ = q v‾×B‾….(i)
Also,
F = ma
Using equation (i) we can write:
ma=qv‾ × B‾
a=qvB/m
Time taken by the bullet to travel 100 m horizontally,
t=d/v
=(100/270)s
Deflection due to the magnetic field in this time interval,
= 3.7 × 10 -6 m.
Question-6 :- (Magnetic Field Exercise HC Verma)
When a proton is released from rest in a room, it starts with an initial acceleration a0towards west. When it is projected towards north with a speed v0, it moves with an initial acceleration 3a0 towards west. Find the electric field and the maximum possible magnetic field in the room.
Answer-6 :-
The initial acceleration of a proton, when it is released from rest, is a0 towards west.
F = qE ….(i)
F = ma0 ….(ii)
Here, q is the charge, E is the electric field and m is the mass.
On equating both the forces F of equations (i) and (ii),
we get:
qE = ma0
⇒ E=ma0/q
towards west
When the proton is projected towards north with a speed v0,
it moves with an initial acceleration 3a0 towards west.
Again, an electric force will act on the proton in the west direction, due to which, an acceleration a0 will act on the proton. Now, as the proton was initially moving with a velocity v, a magnetic force is also acting on the proton.
So, the change in acceleration of the proton will be solely due to the magnetic force acting on it.
Change in acceleration towards west due to the magnetic force
= 3a0 − a0 = 2a0
So,
F = m2a0
Therefore,
required magnetic field,
Question-7 :-
Consider a 10-cm long portion of a straight wire carrying a current of 10 A placed in a magnetic field of 0.1 T making an angle of 53° with the wire. What magnetic force does the wire experience?
Answer-7 :-
A straight wire of length,
l = 10 cm
Electric current flowing through the wire,
I = 10 A
Magnetic field,
B = 0.1 T
Angle between the wire and magnetic field,
θ = 53˚
Magnetic force,
⇒ F = 10 × 10 × 10^−2 ×0.1 ×0.798
⇒F‾ = 0.07989 ≈ 0.08N
The direction of force can be found using Fleming’s left-hand rule.
Therefore, the direction of magnetic force is perpendicular to the wire as well as the magnetic field.
Question-8 :- (Magnetic Field Exercise HC Verma)
A current of 2 A enters at the corner d of a square frame a b c d of side 20 cm and leaves at the opposite corner b. A magnetic field B = 0.1 T exists in the space in a direction perpendicular to the plane of the frame, as shown in the figure (figure 34.E1). Find the magnitude and direction of the magnetic forces on the four sides of the frame.
Answer-8 :-
A square frame abcd of side,
l = 20 cm
Electric current through the wire,
I = 2 A
Magnetic field,
B = 0.1 T
The direction of magnetic field is perpendicular to the plane of the frame, coming out of the plane.
As per the question, current enters at the corner d of the square frame and leaves at the opposite corner b.
Angle between the frame and magnetic field,
θ = 90˚
Magnetic force,
= 0.02 N
The direction of force can be found using Fleming’s left-hand rule.
Thus,
the direction of magnetic force is towards the left.
For wires dc and ab,
The direction of force can be found using Fleming’s left-hand rule.
Thus, the direction of magnetic force is downwards.
Magnetic Field Exercise HC Verma Questions Solutions
(HC Verma Ch-34 Concept of Physics Vol-2 for ISC Class-12)
(Page-231)
Question-9 :-
A magnetic field of strength 1.0 T is produced by a strong electromagnet in a cylindrical region of radius 4.0 cm, as shown in the figure. A wire, carrying a current of 2.0 A, is placed perpendicular to and intersecting the axis of the cylindrical region. Find the magnitude of the force acting on the wire.
Answer-9 :-
Magnetic field,
(B) = 1 T
Radius of the cylindrical region, r = 4.0 cm
Electric current through the wire, I = 2 A
The direction of magnetic field is perpendicular to the plane of the wire.
So, angle between wire and magnetic field,
θ = 90˚
Magnetic force,
= 2 × 8 ×10-2 × 1.0 × 1
= 0.16N
Question-10 :-
A wire of length l carries a current i long the x-axis. A magnetic field exists, which is given as B‾=B0(i‾+j‾+k‾) T. Find the magnitude of the magnetic force acting on the wire.
Answer-10 :-
A wire of length l cm
Electric current through the wire
= i
Magnetic field,
B‾=B0(i‾+j‾+k‾) T
As per the question, the current is passing along the X-axis.
Magnetic force,
F‾=i‾l×B‾
Putting the repective values in the above equation,
we get:
Hence,
|vec F| = sqrt(2ilB_0
Question-11 :-
A current of 5.0 A exists in the circuit shown in the figure. The wire PQ has a length of 50 cm and the magnetic field in which it is immersed has a magnitude of 0.20 T. Find the magnetic force acting on the wire PQ.
Answer-11 :-
Length of the wire PQ inside the magnetic field,
l = 50 cm
Electric current through the wire,
I = 5 A
Magnetic field,
B = 0.2 T
The direction of magnetic field is perpendicular to the plane of the frame and it is going into the plane of the circuit.
As per the question,
Angle between the plane of the wire and the magnetic field,
θ = 90˚
Magnetic force,
= 5 × 50 × 10−2 × 0.2 × 1
= 0.50 N
The direction of force can be found using Fleming’s left-hand rule.
Thus,
the direction of magnetic force is upwards in the plane of the paper.
Question-12 :- (Magnetic Field Exercise HC Verma)
A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field (figure 34-E4). The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire.
Answer-12 :-
A circular loop of radius = a
So, the length of the loop,
l = 2πa
Electric current through the loop = i
As per the question,
The loop is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field. The strength of the magnetic field at the periphery of the loop is B
Therefore, the magnetic field points radially outwards.
Here, the angle between the length of the loop and the magnetic field,
θ = 90˚
Magnetic force is given by
Direction of the force can be found using Fleming’s left-hand rule.
Thus, the direction of magnetic force is perpendicular to the plane of the figure and pointing inside.
Question-13 :-
A hypothetical magnetic field existing in a region is given by B‾=Bo e‾ r where e‾ r denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the x−y plane and the centre at (0, 0, d). Find the magnitude of the magnetic force acting on the loop.
Answer-13 :-
A hypothetical magnetic field existing in a region, B‾=B0e‾ r where denotes the unit vector along the radial direction.
A circular loop of radius a
So, the length of the loop, l = 2πa
Electric current through loop = i
As per the question, the loop is placed with its plane parallel to the X−Y plane and its centre is at (0, 0, d).
Here, angle between the length of the loop and the magnetic field is θ.
Magnetic force is given by
Question-14 :- (Magnetic Field Exercise HC Verma)
A rectangular wire-loop of width a is suspended from the insulated pan of a spring balance, as shown in the figure. A current i exists in the anti-clockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.
Answer-14 :-
A rectangular wire loop of width a
Electric current through the loop = i
Direction of the current is anti-clockwise.
Strength of the magnetic field in the lower region = B
Direction of the magnetic field is into the plane of the loop.
Here, angle between the length of the loop and magnetic field, θ = 90˚
Magnetic force is given by
F‾=ia‾×B‾
The magnetic force will act only on side AD and BC.
As side AD is outside the magnetic field, so F = 0
Magnetic force on side BC is
F‾=ia‾×B‾
= iaBsinθ
= iaB
Direction of force can be found using Fleming’s left-hand rule.
Thus, the direction of the magnetic force is upward.
Similarly if we change the direction of current to clockwise,
the force along BC,
F‾=ia‾×B‾
Thus, the change in force is equal to the change in tension
= iaB − (− iaB) = 2iaB.
Question-15 :-
A current loop of arbitrary shape lies in a uniform magnetic field B. Show that the net magnetic force acting on the loop is zero.
Uniform magnetic field existing in the region of the arbitrary loop
= B
Let the electric current flowing through the loop be i.
Length of each side of the loop is l.
Assume that the direction of the current is clockwise.
Direction of the magnetic field is going into the plane of the loop.
Magnetic force is given by
F‾=il‾×B‾
F‾=i l B sinθ
Here, θ = 90˚
Direction of force can be found using Fleming’s lef- hand rule.
Force F1 acting on AB = ilB upwards
Force F2 acting on DC = ilB downwards
So,
F1 and F2 cancel each other.
Force F3 acting on AD = ilB outwards (Pointing towards the left from AB)
Force F4 acting on BC = ilB outwards (Pointing towards the right from BC)
So,
F3 and F4 cancel each other.
Therefore,
the net force acting on the arbitrary loop is 0.
Question-16 :- (Magnetic Field Exercise HC Verma)
Prove that the force acting on a current-carrying wire, joining two fixed points a and b in a uniform magnetic field, is independent of the shape of the wire.
Uniform magnetic field existing in the region of the wire = B
The electric current flowing through the wire be i.
Length of the wire between two points a and b = l
Magnetic force is given by
F‾=il‾×B‾
θF‾=ilBsinθ
Let us consider two wires of length l, one straight and the other circular.
The circular wire is of radius a such that
2πa=l
Suppose the magnetic field is pointing along the z direction and both the wires are lying in the xy plane, so that the angle between the area vector and the magnetic field is 90°.
For the straight wire of length l lying in a uniform magnetic field of strength B :
Force ,
F=ilBsin(90∘)=ilB
For the circular wire :
Length ,
l=2πa
Angel between the area vector and magnetic field will again be 90°.
Force acting on the circular wire,
F=i(2πa)Bsin(90∘)
= i2πaB = ilB
Both the forces are equal in magnitude. This implies that the magnetic force is independent of the shape of the wire and depends on the length and orientation of the wire.
Therefore,
the magnetic force is independent of the shape of the wire.
Question-17 :-
A semicircular wire of radius 5.0 cm carries a current of 5.0 A. A magnetic field B of magnitude 0.50 T exists along the perpendicular to the plane of the wire. Find the magnitude of the magnetic force acting on the wire.
Answer-17 :-
Radius of semicircular wire,
r = 5.0 cm
Thus, the length of the wire = 2r
Electric current flowing through wire = 5.0 A
Magnetic field,
B = 0.50 T
Direction of magnetic field is perpendicular to the plane of the wire which implies that angle between length of the wire and magnetic field,
θ = 90˚
As we know the magnetic force is given by
F‾= Il‾×B‾
F‾ = I2rBsin90∘
= 5 × 2 × 0.5 × =0.25 N
Question-18 :-
A wire, carrying a current i, is kept in the x−y plane along the curve y = A sin (2xλ/x). magnetic field B exists in the z direction. Find the magnitude of the magnetic force on the portion of the wire between x = 0 and x = λ.
Answer-18 :-
Electric current flowing through the wire = i
The wire is kept in the x−y plane along the curve,
y=A sin ((2x/λ)x)
Magnetic field (B) exists in the z direction.
We have to find the magnetic force on the portion of the wire between x = 0 and x = λ.
Magnetic force is given by
F→=il→×B→
For a small element dl,
dF→=i(dl×B→)
The effective force on the whole wire is equivalent to the force on a starlight wire of length λ placed along the x axis.
So,
Question-19 :- (Magnetic Field Exercise HC Verma)
A rigid wire consists of a semi-circular portion of radius R and two straight sections (figure). The wire is partially immersed in a perpendicular magnetic field B, as shown in the figure. Find the magnetic force on the wire if it carries a current i.
Answer-19 :-
Radius of the semi-circular portion of the rigid wire = R
Magnetic field = B
Electric current flowing through the wire = i
As per the question,
the wire is partially immersed in a perpendicular magnetic field.
As PQ and RS are straight wires of length l each and strength of the magnetic field is also same on both the wires, the force acting on these wires will be equal in magnitude but their directions will be opposite to each other.(Direction of force can be found out using Fleming’s left hand rule.)
So,
the magnetic force on the wire PQ and the force on the wire RS are equal and opposite to each other. Both the forces cancel each other.
Therefore,
only the semicircular loop PR will experience a net magnetic force.
Here,
angle between the length of the wire and magnetic field,
θ = 90˚
Magnetic force in the loop PR,
Question-20 :-
A straight horizontal wire of mass 10 mg and length 1.0 m carries a current of 2.0 A. What minimum magnetic field B should be applied in the region, so that the magnetic force on the wire may balance its weight?
Answer-20 :-
Mass of the wire,
M = 10 mg = 10−5 Kg
Length of the wire,
l = 1.0 m
Electric current flowing through wire,
I = 2.0 A
As per the question, the weight of the wire should be balanced by the magnetic force acting on the wire. Also angle between the length of the wire and magnetic field is 90°
Thus, Mg = IlB, where
g is the acceleration due to gravity = 9.8 m/s2
B is the applied magnetic field
So,
= 4.9 ×10-5 T
Question-21 :- (Magnetic Field Exercise HC Verma)
Figure shows a rod PQ of length 20.0 cm and mass 200 g suspended through a fixed point O by two threads of lengths 20.0 cm each. A magnetic field of strength 0.500 T exists in the vicinity of the wire PQ, as shown in the figure. The wires connecting PQ with the battery are loose and exert no force on PQ. (a) Find the tension in the threads when the switch S is open. (b) A current of 2.0 A is established when the switch S is closed. Find the tension in the threads now.
Answer-21 :-
Length of the rod PQ = 20.0 cm
Mass of the rod,
M = 200 g
Length of the two threads,
l = 20.0 cm
Applied magnetic field,
B = 0.500 T
As per the question,
(a) When the circuit is open:
The weight of the rod is balanced by the tension in the rod. So,
2Tcos30° = Mg
=1.13 N
(b) When the circuit is closed and current flowing through the circuit = 2 A:
Then,
2Tcos 30°= Mg + ilB
= 0.200× 9.8 + 2 × 0.2 × 0.5
= 1.95 + 0.2 = 2.16
⇒ 2T = 2.16×2/√3
⇒ T = 1.245
= 1.25 N
Magnetic Field Exercise HC Verma Questions Solutions
(HC Verma Ch-34 Concept of Physics Vol-2 for ISC Class-12)
(Page-232)
Question-22 :-
Two metal strips, each of length l, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perpendicularly, as shown in the figure. A vertically-upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction between the wire and the floor is µ. A current i is established when the switch S is closed at the instant t = 0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach?
Answer-22 :-
Length of the two metal strips = l
Separation between the strips = b
Mass of the wire = m
Strength magnetic field = B
Coefficient of friction between the wire and the floor = µ
Distance covered by the wire be x.
Due to the presence of the magnetic field, a net magnetic force will act on the wire towards the right.
As the contact between the wire and strip is smooth so coefficient of friction between them is zero. Under the influence of magnetic force,
firstly the wire will travel a distance equal to the length of the strips. After this, it travels a distance x and also now, a frictional force will act on the wire in a direction opposite to its direction of motion.
So we can equate the work done by the magnetic force and the frictional force.
Thus,
F × l = µmg × x,
where g is the acceleration due to gravity
⇒ ibBl = µmgx
⇒ x=iblB/μmg
Question-23 :- (Magnetic Field Exercise HC Verma)
A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 4.90 cm (figure). A vertically-downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0 Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction.
Mass of the metal wire,
M = 10 g
Distance between the two horizontal metal rails,
l = 4.90 cm
Vertically-downward magnetic field,
B = 0.800 T
As per the question, when the resistance of the circuit is slowly decreased below 20.0 Ω, the wire PQ starts sliding on the rails. At that moment,
current in the wire,
i = V/R=(6/20)A
Using Fleming’s left-hand rule, the magnetic force will act towards the right. So, due to this magnetic force, the wire will try to slide on the rails.
When the wire just starts sliding on the rails, the frictional force acting on the wire will just balance the magnetic force acting on the wire. This implies
µR = F, where
µ is the coefficient of friction
R is the normal reaction force and
F is the magnetic force
µ = 0.12
Question-24 :-
A straight wire of length l can slide on two parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is µ. If the wire carries a current i, what minimum magnetic field should exist in the space in order to slide the wire on the rails?
Length of the wire = l
The coefficient of friction between the wire and the rails = µ
Electric current flowing through the wire = i
Distance between the plastic rails = d
The minimum magnetic field that should exist in the space, in order to slide the wire on the rails
µR = F,
where-
µ is the coefficient of friction between the wire and the rail
R is the normal reaction force
F is the magnetic force
µMg = iB
B=μMg/il
Question-25 :-
The figure shows a circular wire loop of radius a and carrying a current i, which is placed in a perpendicular magnetic field B. (a) Consider a small part dl of the wire. Find the force on this part of the wire exerted by the magnetic field. (b) Find the force of compression in the wire.
Radius of the circular wire loop = a
Electric current flowing through the loop = i
Perpendicular magnetic field = B
(a) The force exerted by the magnetic field on a small element dl of the wire,
Fdl = i. (dl × B) = idlB
Using Fleming’s left-hand rule, we can say that the direction of magnetic force is towards the centre for any small element of length dl of the wire (dl and B are perpendicular to each other).
(b) Suppose some part of wire loop subtends a small angle 2θ at the centre of a circular loop
2Tsinθ = i.(dl × B)
= i.2aθ B (Using length of any arc, l =rθ)
if θ is very small, sinθ = θ
2Tθ = i.2aθB
so T = iaB
Question-26 :-
Suppose that the radius of cross-section of the wire used in the previous problem is r. Find the increase in the radius of the loop if the magnetic field is switched off. Young’s modulus of the material of the wire is Y.
Young’s modulus of the material of the wire is Y.
As per the question,
Young’s modulus,
Here,
T is the tension
A is the area of cross-section
Δl is the increase in length of the wire
Question-27 :- (Magnetic Field Exercise HC Verma)
The magnetic field existing in a region is given by B¯=Bo(1+x/1)k¯ . A square loop of edge l and carrying a current i, is placed with its edges parallel to the x−y axes. Find the magnitude of the net magnetic force experienced by the loop.
Answer-27 :-
B¯=Bo(1+x/1)k¯
f1 = force on AB = iB0[1 + 0]l = iBol
f2 = force on CD = iB0[1 + 0]l = iBol
f3 = force on AD = iB0[1 + 0/1]l = iBol
f4 = force on AB = iB0[1 + 1/1]l = 2iBol
Net horizontal force = F1 – F2= 0
Net vertical force = F4 – F3 = iBol
Question-28 :-
A conducting wire of length l, lying normal to a magnetic field B, moves with a velocity v,as shown in the figure. (a) Find the average magnetic force on a free electron of the wire. (b) Due to this magnetic force, electrons concentrate at one end, resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops. (c) What potential difference is developed between the ends of the wire?
Answer-28 :-
Length of the conducting wire = l
Inward magnetic field = B
Velocity of the conducting wire = v
As the wire is moving with velocity v, we can consider this as the net motion of electrons inside the wire with velocity v.
(a) The average magnetic force on a free electron of the wire
= e(v × B) = evB, where e is the charge of an electron.
(b) The redistribution stops when the electric force is just balanced by the magnetic force.
Electric force, F = eE and also magnetic force, F = evB
On aquatinting the two forces, we get:-
eE = evB
⇒ E = vB
(c) The potential difference is developed between the ends of the wire:-
V = lE = lvB, where V is the potential difference across the ends of the wire.
Question-29 :-
A current i is passed through a silver strip of width d and area of cross-section A. The number of free electrons per unit volume is n. (a) Find the drift velocity v of the electrons. (b) If a magnetic field B exists in the region, as shown in the figure, what is the average magnetic force on the free electrons? (c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor, which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons. (d) What will be the potential difference developed across the width of the conductor due to the electron-accumulation? The appearance of a transverse emf, when a current-carrying wire is placed in a magnetic field, is called Hall effect.
Answer-29 :-
Width of the silver strip = d
Area of cross-section = A
Electric current flowing through the strip = i
The number of free electrons per unit volume = n
(a) The relation between the drift velocity and current through any wire,
i = vnAe, where e = charge of an electron and v is the drift velocity.
v=i/nAe
(b)The magnetic field existing in the region is B.
The average magnetic force on a current-carrying conductor,
F = ilB
So, the force on a free electron
=ilb/nAl=iB/nA
upwards (Using Fleming’s left-hand rule)
(c) Let us take the electric field as E.
So, further accumulation of electrons will stop when the electric force is just balanced by the magnetic force.
(d) The potential difference developed across the width of the conductor due to the electron-accumulation will be
Question-30 :- (Magnetic Field Exercise HC Verma)
A particle of charge 2.0 × 10−8 C and mass 2.0 × 10−10 g is projected with a speed of 2.0 × 103 m s−1 in a region with a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.
Charge of the particle, q = 2.0 × 10−8 C
Mass of the particle, m = 2.0 × 10−10 g
Projected speed of the particle, v = 2.0 × 103 m s−1
Uniform magnetic field, B = 0.10 T.
As per the question, the velocity is perpendicular to the field.
So, for the particle to move in a circle,the centrifugal force to the particle will be provided by the magnetic force acting on it.
Using qvB =mv²/r ,
where,
r is the radius of the circle formed,
= 6.28 × 10-4 s
Question-31 :-
A proton describes a circle of radius 1 cm in a magnetic field of strength 0.10 T. What would be the radius of the circle described by an α-particle moving with the same speed in the same magnetic field?
Radius of the circle, r = 1 cm
Magnetic field = 0.10 T
We know that the charge of a proton is e and that of an alpha particle is 2e. Also, the mass of a proton is m and that of an aplha particle is 4m.
Suppose, both the particles are moving with speed v.
According to the question,
rp =mv/eB,
where,
rp is the radius of the circle described by the proton
Question-32 :- (Magnetic Field Exercise HC Verma)
An electron of kinetic energy 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.
Answer-32 :-
Given:
Kinetic energy of an electron = 100 eV
Radius of the circle = 10 cm
(1/2)mv² = 100 eV = 1.6 × 10−17 J (1 eV = 1.6 × 10−19 J)
Here,
m is the mass of an electron and v is the speed of an electron.
Thus,
1/2 × 9.1 × 10−31 × v2 = 1.6 × 10−17 J
⇒ v2 = 0.35 × 1014
v = 0.591 × 107 m/s
Now,
Therefore, the applied magnetic field = 3.4 × 10−4 T
Number of revolutions per second of the electron,
= 9.4 × 106
f = 9.4 × 106
Question-33 :-
Protons with kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field, so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field.
Answer-33 :-
Distance of the target from the accelerator = l
Therefore, radius of the circular orbit ≤ l
As per the question, the beam is bent by a perpendicular magnetic field.
We know,
For a proton, the above equation can be written as:
l =mpv/eB (As r=l)….(i)
Here,
mp is the mass of a proton
v is the velocity
e is the charge
B is the magnetic field
Magnetic Field Exercise HC Verma Questions Solutions
(HC Verma Ch-34 Concept of Physics Vol-2 for ISC Class-12)
(Page-233)
Question-34 :-
A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 106 m s−1. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it.
Answer-34 :-
Applied potential difference,
V = 12 kV = 12 × 103 V
Speed of a charged particle,
v =1.0 × 106 m s−1
Magnetic field strength,
B = 0.2 T
As per the question, a charged particle is injected perpendicularly into the magnetic field.
We know:
Question-35 :- (Magnetic Field Exercise HC Verma)
Doubly-ionized helium ions are projected with a speed of 10 km s−1 in a direction perpendicular to a uniform magnetic field of magnitude 1.0 T. Find (a) the force acting on an ion (b) the radius of the circle in which it circulates and (c) the time taken by an ion to complete the circle.
Answer-35 :-
Speed of the helium ions,
v = 10 km s−1 = 104 m/s
Uniform magnetic field,
B = 1.0 T
Charge on the helium ions = 2e
Mass of the helium ions,
m = 4 × 1.6 × 10-27 kg
(a) The force acting on an ion,
F = qvBsinθ
= 2 × 1.6 × 10−19 × 104 × 1.0
= 3.2 × 10−15 N
(b) The radius of the circle in which it circulates,
(c) The time taken by an ion to complete the circle,
= 1.31 × 10−7 s
Question-36 :-
A proton is projected with a velocity of 3 × 106 m s−1 perpendicular to a uniform magnetic field of 0.6 T. Find the acceleration of the proton.
Uniform magnetic field,
As per the question, the proton is projected perpendicular to a uniform magnetic field.
We know,
F = mpa ….(i)
and
F = evBsinθ ….(ii)
On equating (i) and (ii), we get:
ma = evBsinθ (As θ = 90˚)
= 1.72 × 1014 m/s2
Question-37 :-
(a) An electron moves along a circle of radius 1 m in a perpendicular magnetic field of strength 0.50 T. What would be its speed? Is it reasonable?
(b) If a proton moves along a circle of the same radius in the same magnetic field, what would be its speed?
(a) Radius of the circle = 1 m
Magnetic field strength = 0.50 T
We know:
r=meve/Be
ve=rBe/BE,
where/ me is mass of the electron of the speed of the electron
≈ 8.8 × 1010 m/s
Since, the speed of the electron moving along the circle is greater than the speed of light, it is not reasonable.
(b) For a proton,
r=mevp/Be
vp=rBe/mp
where mp is the mass of the proton and vp is its speed.
r = 5 × 107 m/s
Question-38 :-
A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B, as shown in the figure. (a) Find the radius of the circular arc it describes in the magnetic field. (b) Find the angle subtended by the arc at the centre. (c) How long does the particle stay inside the magnetic field? (d) Solve the three parts of the above problem if the charge q on the particle is negative.
Answer-38 :-
Mass of the particle = m
Positive charge on the particle = q
Uniform velocity of the particle = v
Magnetic field = B
(a) The radius of the circular arc described by the particle in the magnetic field:-
We know
r=mv/qB
(b)
The angle subtended by the arc at the centre:-
Line MAB is tangent to arc ABC, so the angle described by the charged particle,
∠MAO = 90°
Now, ∠NAC = 90°
OAC = OCA = θ
[by geometry]
Then, AOC = 180° − (θ + θ) (By angle-sum property of a triangle)
= π − 2θ
(c) The time for which the particle stay inside the magnetic field:-
Distance covered by the particle inside the magnetic field,
l = rθ
(d) If the charge q on the particle is negative, then
(i) Radius of circular arc,
r=mv/qB
(ii) The centre of the arc will lie within the magnetic field. Therefore, the angle subtended by the arc = π + 2θ
(iii) Similarly, the time taken by the particle to cover the path inside the magnetic field =
Question-39 :- (Magnetic Field Exercise HC Verma)
A particle of mass m and charge q is projected into a region that has a perpendicular magnetic field B. Find the angle of deviation (figure 34 -E14) of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than
Answer-39 :-
Mass of the particle = m
Charge of the particle = q
Magnetic field = B
As per the question, the particle is projected into a perpendicular magnetic field.
(a) When the width,
d = mv/qB
d is equal to the radius and θ is the angle between the radius and tangent, which is equal to π/2.
(b) When the width,
d = mv/2qB
Width of the region in which a magnetic field is applied is half of the radius of the circular path described by the particle.
As the magnetic force is acting only along the y direction, the velocity of the particle will remain constant along the x direction. So, if d is the distance travelled along the x axis, then
d = vxt
t=d/Vx………(i)
(i)
(ii)
The acceleration along the x direction is zero. The force will act only along the y direction.
Using the equation of motion for motion along the y axis:
vy = uy + ayt
Putting the value of t from equation (i),
we get:
(c) When the width, d = (c)
When the width,
d = 2mv/qB
Question-40 :-
A narrow beam of singly-charged carbon ions, moving at a constant velocity of 6.0 × 104m s−1, is sent perpendicularly in a rectangular region of uniform magnetic field B = 0.5 T (figure 34 -E 15). It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion = A(1.6 × 10−27) kg, where A is the mass number.
(figure 34 -E 15)
Answer-40 :-
Velocity of a narrow beam of singly-charged carbon ions,
v = 6.0 × 104 m s−1
Strength of magnetic field
B = 0.5 T
Separations between the two beams from the incident beam are 3.0 cm and 3.5 cm.
Mass of an ion
= A(1.6 × 10−27) kg
The radius of the curved path taken by the first beam,
r1 = m1v/qB
where m1 is the mass of the first isotope and q is the charge.
For the second beam:
r2=m2v/qB
where m2 is the mass of the first isotope and q is the charge.
= 14.58 u
So, the two isotopes of carbon used are 12C6 and 14C6.
Question-41 :-
Fe+ ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20.0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A (1.6 × 10−27) kg, where A is the mass number.
V = 500V B = 20mT = (2 × 10−3)T
= 1.19 × 10−2 m
= 119 cm
= 1.20 × 10−2 m
= 120 cm
Question-42 :- (Magnetic Field Exercise HC Verma)
A narrow beam of singly charged potassium ions of kinetic energy 32 keV is injected into a region of width 1.00 cm with a magnetic field of strength 0.500 T, as shown in the figure. The ions are collected at a screen 95.5 cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion = A (1.6 × 10−27) kg, where A is the mass number.
Answer-42 :-
Kinetic energy of singly-charged potassium ions = 32 keV
Width of the magnetic region = 1.00 cm
Magnetic field’s strength,
B = 0.500 T
Distance between the screen and the region = 95.5 cm
Atomic weights of the two isotopes are 39 and 41.
Mass of a potassium ion = A (1.6 × 10−27) kg
For a singly-charged potassium ion K-39 :
Mass of
K-39 = 39 × 1.6 × 10−27 kg,
Charge,
q = 1.6 × 10−19 C
As per the question, the narrow beam of singly-charged potassium ions is injected into a region of magnetic field.
As
K.E = 32 keV
1/2mv² = 32×10³×1.6×10−19
(1/2)× 39 × (1.6 × 10−27) × v2 = 32 × 103 × 1.6 × 10−19
v = 4.05 × 105
We know that throughout the motion, the horizontal velocity remains constant.
So, the time taken to cross the magnetic field,
t = (d/v)
=0.01/4.05×105
= 24.7 × 10−9 s
Now, the acceleration in the magnetic field region:
F = qvB = ma
= 5192 × 108 m/s2
Velocity in the vertical direction,
vy = at
= 5193.53 × 108 × 24.7 × 10−9
= 12824.24 m/s
Time taken to reach the screen
(d/v) = 0.955/4.05×105
Distance moved vertically in this time
= vy × t
= 12824.24 × 2358 × 10−9
= 3023.95×10-5 m
Vertical distance travelled by the particle inside magnetic field can be found out by using equation of motion
v2 = 2aS
⇒ (12824.24)2
= 2 × 5192 × 108 × S
⇒ 15.83×10−5 = S
Net display from line
= 15.83×10−5+3023.95×10-5
= 3039.787×10−5 m.
12× 41 × 1.6 × 10−27 v2
= 32 × 103 × 1.6 × 10−9
⇒ v = 3.94×105 m/s
Similarly, acceleration,
a = 4805 × 108 m/s2
t = time taken for exiting the magnetic field
= 25.4 × 10−9 sec.
vy1= at (vertical velocity)
= 4805 × 108 × 25.4 × 10−9
= 12204.7× 10−9 m/s
Time to reach the screen
= 2423× 10−9 s.
Distance moved vertically
= 12204.7 × 2423 × 10−9
= 2957.1× 10−5
Now,
Vertical distance travelled by the particle inside magnetic field can be found out by using equation of motion
v2 = 2aS
(12204.7)2
= 2 × 4805 × 108 S
⇒ S = 15.49× 10-5 m
Net distance travelled
= 15.49× 10-5 +2957.1× 10−5
= 2972.68× 10-5 m.
Net gap between K-39 and K-41
= 3039.787×10−5− 2972.68× 10-5
= 67 mm.
Question-43 :-
The figure shows a convex lens of focal length 12 cm lying in a uniform magnetic field Bof magnitude 1.2 T parallel to its principal axis. A particle with charge 2.0 × 10−3 C and mass 2.0 × 10−5 kg is projected perpendicular to the plane of the diagram with a speed of 4.8 m s−1. The particle moves along a circle with its centre on the principal axis at a distance of 18 cm from the lens. Show that the image of the particle moves along a circle and find the radius of that circle.
Answer-43 :-
Focal length of the convex lens = 12 cm
Uniform magnetic field,
B = 1.2 T
Charge of the particle,
q = 2.0 × 10−3 C
and mass,
m = 2.0 × 10−5 kg
Speed of the particle,
v = 4.8 m s−1
The distance of the particle from the lens = 18 cm
As per the question, the object is projected perpendicular to the plane of the paper.
Let the radius of the circle on which the object is moving be r.
We know:
Here, object distance,
u = -18 cm
Using the lens equation
Image distance,
v = 36 cm.
The radius of the circular path of image be r’.
So, magnification:
= 8 cm
Therefore, the radius of the circular path in which the image moves is 8 cm.
Question-44 :- (Magnetic Field Exercise HC Verma)
Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference V along the x-axis. These electrons emerge from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles, as shown in the figure. Show that these paraxial electrons are refocused on the x-axis at a distance 
Answer-44 :-
Electrons are accelerated through a potential difference = V
The mass of an electron be m and the charge of an electron be e.
Electric field,
E = Vr
Force experienced by the electron,
F = eE
Acceleration of the electron,
a = eV/rm
Using the equation of motion
v2 − u2= 2 × a × s,
v2 = 2 × a × s (As u = 0)
Here, s = r
Time taken by electron to cover the curved path,
As the acceleration of the electron is along the y axis only, it travels along the x axis with uniform velocity.
Velocity of the electron moving along the field remains v.
Therefore, the distance at which the beam is refocused,
d = v × T
Magnetic Field Exercise HC Verma Questions Solutions
(HC Verma Ch-34 Concept of Physics Vol-2 for ISC Class-12)
(Page-234)
Question-45 :-
Two particles, each with mass m are placed at a separation d in a uniform magnetic field B, as shown in the figure. They have opposite charges of equal magnitude q. At time t = 0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off. (a) Find the maximum value vmof the projection speed, so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if v = vm/2? (c) At what instant will a collision occur between the particles if v = 2vm? (d) Suppose v = 2vm and the collision between the particles is completely inelastic. Describe the motion after the collision.
Answer-45 :-
Mass of two particles = m
Distance between them = d
Both the particles have equal charges in magnitude but opposite polarity equal to q.
As per the question, both the particles are projected towards each other with equal speed v.
It is assumed that Coulomb force between the charges is switched off.
(a) The maximum value vm of the projection speed so that the two particles do not collide:-
Both the particles will not collide if
d = r1 + r2
(where, r1 = r2 = radius of circular orbit described by the charged particles)
(b) The minimum and maximum separation between the particles if v = vm/2:-
Let the radius of the curved path taken by the particles, when they are projected with speed vm/2, be r.
So, minimum separation between the particles = (d – 2r)
Maximum distance separation = (d + 2r)
(c) The instant at which the collision occurs between the particles when
The particles will collide at a distance d/2 along the horizontal direction.
They collide after time t.
Velocity of the particles along the horizontal direction will remain the same.
(d) The motion of the two particles after collision when the collision is completely inelastic:-
v = 2vm
Let the particles collide at point P.
And at point P, both the particles will have motion in upward direction.
As the collision is inelastic they stick together.
Distance between centers = d
Velocity along the horizontal direction does not get affected due to the magnetic force.
At point P, velocities along the horizontal direction are equal and opposite. So, they cancel each other.
Velocity along the vertical direction (upward) will add up.
Magnetic force acting along the vertical direction,
F=q(2vm)B
Acceleration along the vertical direction,
Velocity of the combined mass at point P is along the vertical direction.
So,
Hence,
both the particles will behave as a combined mass and move with velocity vm.
Question-46 :-
Mass of the particle,
Charge of the particle,
As per the question, if the particle has to move with uniform velocity in the region of the applied field,
gravitational force experienced by the particle should be equal to the magnetic force experienced by the particle.
So, qvB = mg,
⇒ 1 × 10−5 × v × 2 × 10−1
⇒ v = 4.9 × 10
Question-47 :-(Magnetic Field Exercise HC Verma)
A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V m−1 makes the path straight. Find the charge/mass ratio of the particle.
Diameter of the circle = 1.0 cm
Thus, radius of circle,
r = = 0.5 × 10−2 m,
Magnetic field,
B = 0.40 T
Electric field,
E = 200 V m−1.
As per the question, the particle is moving in a circle under the action of a magnetic field. But when an electric field is applied on the particle, it moves in a straight line.
So, we can write:
Fe = Fm
qE = qvB,
where q is the charge and v is the velocity of the particle.
Question-48 :-
A proton goes unelected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of 2.0 × 105 m s−1. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and magnetic fields. Take the mass of the proton = 1.6 × 10−27 kg
Mass of the proton,
m = 1.6 × 10−27 kg
Speed of the proton inside the crossed electric and magnetic field,
v = 2.0 × 105 ms−1
As per the question, the proton is not deflected under the combined action of the electric and magnetic fields. Thus, the forces applied by both the fields are equal and opposite.
That is,
qE = qvB
⇒ E = vB …(1)
But when the electric field is switched off, the proton moves in a circle due to the force of the magnetic field.
Radius of the circle,
r = 4.0 cm
= 4 × 10−2 m
= 0.5 × 10−1 = 0.05 T
Putting the value of B in equation (1),
we get:
E = 2 × 105 × 0.05
= 1 × 104 N/c
Question-49 :-
A particle with a charge of 5.0 µC and a mass of 5.0 × 10−12 kg is projected with a speed of 1.0 km s−1 in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin−1 (0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.
Answer-49 :-
Charge of the particle,
q = 5 µC = 5 × 10−6 C
Magnetic field intensity,
B = 5 × 10−3 T
Mass of the particle,
m = 5 × 10−12 kg
Velocity of projection,
v = 1 Km/s
= 103 m/s
Angle between the magnetic field and velocity,
θ= sin−1(0.9)
Component of velocity perpendicular to the magnetic field,
v⊥= v sinθ
Component of velocity in the direction of magnetic field, v∣∣
Since there are no forces in the horizontal direction (the direction of magnetic field), the particle moves with uniform velocity.
The velocity has a vertical component along which it accelerates with an acceleration a and moves in a circular cross-section.
Thus,
It moves in a helix.
Hence, diameter of the helix, 2r = 0.36 m = 36 cm
Pitch,
= 0.55 m
= 55cm
Question-50 :-
A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton = 1.6 × 10−27 kg
Magnetic field intensity, B = 0.02 T
Radius of the helical path, r = 5 cm = 5 × 10−2 m
Pitch of the helical path, p = 20cm = 2 × 10−1 m
We know that for a helical path, the velocity of the proton has two components,
= 6.4 × 10^4 m/s
Question-51 :-
A particle of mass m and charge q is released from the origin in a region in which the electric field and magnetic field are given by
Find the speed of the particle as a function of its z-coordinate.
Mass of the particle = m
Charge of the particle = q
Electric field and magnetic field are given by
Here,
z is the distance along the z-direction.
Question-52 :- (Magnetic Field Exercise HC Verma)
An electron is emitted with negligible speed from the negative plate of a parallel-plate capacitor charged to a potential difference V. The separation between the plates is d and a magnetic field B exists in the space, as shown in the figure. Show that the electron will fail to strike the upper plates if
(figure 34-E20)
Answer-52 :-
Potential difference across the plates of the capacitor = V
Separation between the plates = d
Magnetic field intensity = B
The electric field set up between the plates of a capacitor,
E=V/d
The electron will move in a circular path due to the given magnetic field. Radius of the circular path,
r=meV/eB
And the electron will fail to strike the upper plate only when the radius of the circular path will be less than d,
ex. d > r
Question-53 :- (Magnetic Field Exercise HC Verma)
A rectangular coil of 100 turns has length 5 cm and width 4 cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2 A is sent through the coil. Find the magnitude of the magnetic field B if the torque acting on the coil is 0.2 N m−1
No. of turns in the coil,
n = 100
Area of the coil,
A = 5 × 4 cm2 = 20 × 10−4 m2
Magnitude of current = 2 A
Torque acting on the coil,
τ = 0.2 N m−1
τ = niA × B
⇒τ = niBA sin 90°
⇒ 0.2 = 100 × 2 ×20 × 10−4 × B
⇒ B = 0.5 T
Question-54 :-
A 50-turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?
No. of turns of the coil,
n = 50
Magnetic field intensity,
B = 0.20 T = 2 × 10−1 T
Radius of the coil,
r = 0.02 m = 2 × 10−2 m
Magnitude of current =5 A
Torque acting on the coil,
τ = niABsinθ
Here,
A is the area of the coil and θ is the angle between the area vector and the magnetic field.
τ is maximum when
θ = 90°.
τmax = niABsin90°
= 50 × 5 × 3.14 × 4 × 10−4 × 2 × 10−1
= 6.28 × 10−2 N-m
τ=12×τmax
⇒ sinθ=12
⇒ θ = 30°
So,
the angle between the magnetic field and the plane of the coil
= 90° − 30° = 60°
Question-55 :-
A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A. A uniform magnetic field of magnitude 0.20 T exists parallel to the longer side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop?
l = 20cm = 20 × 10−2m
B = 10cm = 10 × 10−2m
i = 5A, B = 0.2T
(a) There is no force on the sides AB and CD. But the force on the sides AD and BC are opposite. So they cancel each other.
(b) Torque on the loop
τ = ni vector A x vector B = niABSin90°
= 1 × 5 × 20 × 10−2 × 10× 10−2 0.2 = 2 × 10−2 = 0.02N-M
Parallel to the shorter side.
Question-56 :-
A circular coil of radius 2.0 cm has 500 turns and carries a current of 1.0 A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0.40 T that exists in the space. Find the torque acting on the coil.
Answer-56 :-
No. of turns of the coil,
n = 500
Magnetic field intensity,
B = 0.40 T = 4 × 10−1 T
Radius of the coil,
r = 2 cm
= 2 × 10−2 m
Magnitude of current, i = 1 A
Angle between the area vector and magnetic field, θ = 30°
Torque acting on the coil,
τ = niABsinθ
Here,
A is the area of the coil.
τ = 500 × 1 × 3.14 × 4 × 10−4 ×4×10−1 × 1/2
= 12.56 × 10−2
= 0.1256
= 0.13 N-m
Question-57:-
A circular loop carrying a current i is made of a wire of length L. A uniform magnetic field B exists parallel to the plane of the loop. (a) Find the torque on the loop. (b) If the same length of the wire is used to form a square loop, what would be the torque? Which is larger?
Answer-57 :-
Given,
Magnetic field intensity = B
Circumference,
L = 2πr,
where r is the radius of the coil.
So, area of the coil,
A =l²/4π
Magnitude of current = i
Torque acting on the coil,
τ = niABsinθ,
where θ is the angle between the area vector and the magnetic field.
So,
the torque on the circular loop is larger.
Question-58:- (Magnetic Field Exercise HC Verma)
A square coil of edge l and with n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over?
Answer-58 :-
Number of turns in the coil = n
Edge of the square loop = l
Magnetic field intensity = B
Magnitude of current = i
Angle between area vector and magnetic field,
θ = 90°
Torque acting on the coil due to magnetic field,
τ = ni AB sinθ
Here, A is the area of the coil.
τ=nifBsin90∘
Torque produced due to weight, τweight =
mgl/2
For the coil to start tipping over,
τ ≥ τweight
For minimum value of B,
τ = τweight
Question-59:-
Consider a non-conducting ring of radius r and mass m that has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed ω. (a) Find the equivalent electric current in the ring. (b) Find the magnetic moment µ of the ring. (c) Show that π=q/2m l,
where l is the angular momentum of the ring about its axis of rotation.
Answer-59 :-
Radius of the ring = r
Mass of the ring = m
Total charge of the ring = q
(b) For a ring of area A with current i, magnetic moment,
c) Angular momentum, l = Iω
where I is moment of inertia of the ring about its axis of rotation.
I =mr2
so ,
I =mr2ω
⇒ ωr2=1/m
Putting this value in equation (i), we get:
μ=ql/2m
Magnetic Field Exercise HC Verma Questions Solutions
(HC Verma Ch-34 Concept of Physics Vol-2 for ISC Class-12)
(Page-235)
Question-60:-
Consider a non-conducting plate of radius r and mass m that has a charge q distributed uniformly over it. The plate is rotated about its axis with an angular speed ω. Show that the magnetic moment µ and the angular momentum l of the plate are related as μ = (q/2m)l
Answer-60 :-
Radius of the ring = r
Mass of the ring = m
Total charge of the ring = q
For the ring of area A with current i, magnetic moment,
Angular momentum,
l= Iw,
where I is the moment of inertia of the ring about its axis of rotation.
Putting this value in equation (i),
μ=ql/2m
Question-61:-
Consider a solid sphere of radius r and mass m that has a charge q distributed uniformly over its volume. The sphere is rotated about its diameter with an angular speed ω. Show that the magnetic moment µ and the angular momentum l of the sphere are related as μ = (q/2m)l
Answer-61 :-
Considering the strip of width dx at a distance x from the centre of the sphere.
Small area of the strip is given as,
da = 4π x dx
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