Magnetic Field Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-34

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Magnetic Field Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-34 Concept of Physics for Class-12. Step by Step Solutions of Objective -1 (MCQ-1) Questions of Chapter-34 Magnetic Field (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Magnetic Field Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-34

Board ISC and other board
Publications Bharti Bhawan Publishers
Chapter-34 Magnetic Field 
Class 12
Vol  2nd
writer HC Verma
Book Name Concept of Physics
Topics Solution of Objective-1 (MCQ-1) Questions
Page-Number 229

-: Select Topics :-

Question for Short Answer

Obj-1

Obj-2

Exercise


Magnetic Field Obj-1 (MCQ-1) Questions

HC Verma Solutions of Ch-34 Vol-2 Concept of Physics for Class-12

(Page-229)

Question-1 :-

A positively-charged particle projected towards east is deflected towards north by a magnetic field. The field may be

(a) towards west

(b) towards south

(c) upwards

(d) downwards

Answer 1 :-

The option (d) downwards is correct
Explanation:
A positively-charged particle projected towards east can be considered as current in the eastern direction. Here, the positive charge is deflected towards the north by a magnetic field, i.e. the positively-charged particle experiences a force in the northern direction.

Question-2 :-

A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one point. If a magnetic field is switched on in the vertical direction, the tension in the string

(a) will increase

(b) will decrease

(c) will remain the same

(d) may increase or decrease

Answer 2 :-

The option (d) may increase or decrease is correct
Explanation:
A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one point. If a magnetic field is switched on in the vertical direction, the tension in the string

When the charged particle is whirled in a horizontal circle, at any moment, the current direction can be taken along the tangent of the circle. Also, the magnetic field is in the vertical direction.

Question-3 :-

Which of the following particles will experience maximum magnetic force (magnitude) when projected with the same velocity perpendicular to a magnetic field?

(a) Electron

(b) Proton

(c) He+

(d)  Li++

Answer 3 :-

The option (d)  Li++ is correct
Explanation:

Force on a moving charged particle,
F = qV B sin θ
​As velocity V and magnetic field are constant,  and angle between the magnetic field and charged particle is 90° , the only thing on which force F depends is charge q.
Now, the charge on Li++  charge on electron, proton or He+ . So the force is maximum for Li++.

Question-4 :-

Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field?

(a) Electron

(b) Proton

(c) He+

(d) Li+

Answer 4 :-

The option (a) Electron is correct
Explanation:

When a particle moves in a magnetic field, the necessary centripetal force, for  the particle to move in a circle, is provided by the magnetic force acting on the particle.

Question-5 :-

Which of the following particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?

(a) Electron

(b) Proton

(c) He+

(d) Li+

Answer 5 :-

The option (d) Li+ is correct
Explanation:

Time period of the revolution of the particle,

T = 2πm/qB
As frequency is the reciprocal of time period, so

 f = qB/2πm
The charge on all the four particles is same. But the mass is maximum for Li+. So, it will have the smallest frequency of revolution.

Question-6 :-

A circular loop of area 1 cm2, carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is

(a) zero

(b) 10−4 N m

(c) 10−2 N m

(d) 1 N m

Answer 6 :-

The option (a) zero is correct
Explanation:

When a circular loop is placed in a uniform magnetic field, it always experiences zero toque. We all know that a current-carrying wire experiences a force when placed in an external magnetic field. But in the case of a circular loop, forces are present in pairs, i.e. they are equal and opposite in magnitude. So, for every point on the loop, there exists another point on the diametrically opposite edge for which the force is equal and opposite to the force  acting on first point. So, these two forces cancel in pair. In this way, the net torque on the loop is always zero when placed in a uniform magnetic field.

Question-8 :-

A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be

(a) a straight line

(b) a circle

(c) a helix with uniform pitch

(d) a helix with non-uniform pitch

Answer 8 :-

The option (c) a helix with uniform pitch is correct

Question-9 :-

A particle moves in a region with a uniform magnetic field and a parallel, uniform electric field. At some instant, the velocity of the particle is perpendicular to the field direction. The path of the particle will be

(a) a straight line

(b) a circle

(c) a helix with uniform pitch

(d) a helix with non-uniform pitch

Answer 9 :-

The option (d) a helix with non-uniform pitch is correct
Explanation:

Here, the total Lorentz force on the particle,
F = qE + qVB 
We all know that magnetic field B does not change the speed of the particle but changes its direction. But as an electric field is also present that accelerate the particle in the direction of the field, the resultant path is a helix with a non-uniform pitch.

Question-10 :-

An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged particle q, moving along the axis of the circular wire, passes through its centre at speed v. The magnetic force acting on the particle, when it passes through the centre, has a magnitude equal to

(a) Magnetic Field HC Verma Solutions of Que for Short Ans Ch-34 Vol-2 img 2

(b) Magnetic Field HC Verma Solutions of Que for Short Ans Ch-34 Vol-2 img 3

(c) Magnetic Field HC Verma Solutions of Que for Short Ans Ch-34 Vol-2 img 4

(d) Zero

Answer 10 :-

The option (d) Zero s correct
Explanation:
We can use the right-hand thumb rule to get the direction of magnetic field due to the current-carrying wire. Based on this, it can be determined that the direction of magnetic field is along the axis of the wire. Also, the charged particle is moving along the axis. So, no magnetic force will act on it, as the angle between the magnetic field and the velocity of the charged particle may be   0^° or 180 ° . So, sin θ of the angles between velocity and magnetic field is zero.
Also, the force, F=qVB sin θ
So, the force on the charged particle is zero.

—: End of Magnetic Field Obj-1 HC Verma (MCQ-1)  Solutions Vol-2 Chapter-34 :–


Return to  — HC Verma Solutions Vol-2 Concept of Physics

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