Magnetic Field Exercise HC Verma Solutions Ch-34 Vol-2

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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

-: Select Topics :-

Question for Short Answer

Objective-I

Objective-II

Exercise (currently open)

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.
= 2 × 1.6 × 10−19 C,

Magnetic force acting on the α-particle,

F‾ = qv‾ × B‾ in90°
= 2 × 1.6 × 10−19 × 3 × 10−7
= 9.6 × 10−19 N , towards west

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.

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,

Magnetic Field Exercise HC Verma Solutions Ch-34 Vol-2 img 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.

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,

Magnetic Field Exercise HC Verma Solutions Ch-34 Vol-2 img 4

s = 0.0148 = 1.5 × 10−2 cm

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