Q: In the product \vec{F} = q(\vec{v} \times \vec{B}) = q\vec{v} \times (B\hat{i} + B\hat{j} + B_0\hat{k}). For q = 1 and \vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k} and \vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}, what will be the complete expression for \vec{B}?

-6\hat{i}-6\hat{j}-8\hat{k}. Taking \vec{v} \times \vec{B} with \vec{B} = B\hat{i}+B\hat{j}+B_0\hat{k} and equating components to \vec{F}=4\hat{i}-20\hat{j}+12\hat{k} gives 2B-4B=12 \Rightarrow B=-6, and 6B-2B_0=-20 \Rightarrow B_0=-8. So \vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}.

Q: An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 \times 10^{-2}\ \text{T}. If the value of e/m is 1.76 \times 10^{11}\ \text{C/kg}, the frequency of revolution of the electron is

1\ \text{GHz}. The frequency of revolution is f = \dfrac{1}{2\pi}\left(\dfrac{e}{m}\right)B = \dfrac{1.76\times 10^{11}\times 3.57\times 10^{-2}}{2\pi} \approx 1.0\times 10^{9}\ \text{Hz} = 1\ \text{GHz}.

Q: An alternating electric field of frequency \nu is applied across the dees (radius = R) of a cyclotron that is being used to accelerate protons (mass = m). The operating magnetic field B used in the cyclotron and the kinetic energy K of the proton beam produced by it are given by

B = \dfrac{2\pi m\nu}{e} and K = 2m\pi^2\nu^2R^2. Cyclotron frequency \nu = \dfrac{eB}{2\pi m} \Rightarrow B = \dfrac{2\pi m\nu}{e}. Maximum speed v = 2\pi R \nu, so K = \dfrac{1}{2}mv^2 = 2m\pi^2\nu^2R^2.

Q: The magnetic force acting on a charged particle of charge -2\ \mu\text{C} in a magnetic field of 2\ \text{T} acting in the y-direction, when the particle velocity is (2\hat{i} + 3\hat{j}) \times 10^{6}\ \text{ms}^{-1}, is

8\ \text{N} in -z-direction. \vec{F} = q(\vec{v}\times\vec{B}) = (-2\times10^{-6})\{(2\hat{i}+3\hat{j})\times10^6 \times 2\hat{j}\} = (-2\times10^{-6})[4\times10^6\hat{k}] = -8\ \text{N}\hat{k}, i.e., 8\ \text{N} along the negative z-axis.

Question 1

In the product \vec{F} = q(\vec{v} 	imes \vec{B}) = q\vec{v} 	imes (B\hat{i} + B\hat{j} + B_0\hat{k}). For q = 1 and \vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k} and \vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}, what will be the complete expression for \vec{B}?

Accepted Answer

-6\hat{i}-6\hat{j}-8\hat{k}. Taking \vec{v} 	imes \vec{B} with \vec{B} = B\hat{i}+B\hat{j}+B_0\hat{k} and equating components to \vec{F}=4\hat{i}-20\hat{j}+12\hat{k} gives 2B-4B=12 \Rightarrow B=-6, and 6B-2B_0=-20 \Rightarrow B_0=-8. So \vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}.

Question 2

Ionised hydrogen atoms and \alpha-particles with same momenta enter perpendicular to a constant magnetic field B. The ratio of radii of their paths r_H : r_\alpha will be

Accepted Answer

2:1. Since r = \dfrac{mv}{qB} = \dfrac{p}{qB} and momentum p is the same for both, \dfrac{r_H}{r_\alpha} = \dfrac{q_\alpha}{q_H} = \dfrac{2e}{e} = 2:1.

Question 3

An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 	imes 10^{-2}\ 	ext{T}. If the value of e/m is 1.76 	imes 10^{11}\ 	ext{C/kg}, the frequency of revolution of the electron is

Accepted Answer

1\ 	ext{GHz}. The frequency of revolution is f = \dfrac{1}{2\pi}\left(\dfrac{e}{m}ight)B = \dfrac{1.76	imes 10^{11}	imes 3.57	imes 10^{-2}}{2\pi} \approx 1.0	imes 10^{9}\ 	ext{Hz} = 1\ 	ext{GHz}.

Question 4

An alternating electric field of frequency 
u is applied across the dees (radius = R) of a cyclotron that is being used to accelerate protons (mass = m). The operating magnetic field B used in the cyclotron and the kinetic energy K of the proton beam produced by it are given by

Accepted Answer

B = \dfrac{2\pi m
u}{e} and K = 2m\pi^2
u^2R^2. Cyclotron frequency 
u = \dfrac{eB}{2\pi m} \Rightarrow B = \dfrac{2\pi m
u}{e}. Maximum speed v = 2\pi R 
u, so K = \dfrac{1}{2}mv^2 = 2m\pi^2
u^2R^2.

Question 5

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of the fields, then the electron

Accepted Answer

speed will decrease. Since \vec{v} \parallel \vec{B}, the magnetic force q\vec{v}	imes\vec{B} = 0. Only the electric field acts, exerting a force opposite to the electron's velocity (as the electron is negative), so its speed decreases.

Question 6

A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by

Accepted Answer

\dfrac{E^2}{2VB^2}. For no deflection, v = E/B. Also, eV = \dfrac{1}{2}mv^2 \Rightarrow \dfrac{e}{m} = \dfrac{v^2}{2V} = \dfrac{E^2}{2VB^2}.

Question 7

The magnetic force acting on a charged particle of charge -2\ \mu	ext{C} in a magnetic field of 2\ 	ext{T} acting in the y-direction, when the particle velocity is (2\hat{i} + 3\hat{j}) 	imes 10^{6}\ 	ext{ms}^{-1}, is

Accepted Answer

8\ 	ext{N} in -z-direction. \vec{F} = q(\vec{v}	imes\vec{B}) = (-2	imes10^{-6})\{(2\hat{i}+3\hat{j})	imes10^6 	imes 2\hat{j}\} = (-2	imes10^{-6})[4	imes10^6\hat{k}] = -8\ 	ext{N}\hat{k}, i.e., 8\ 	ext{N} along the negative z-axis.

Question 8

A particle of mass m, charge q and kinetic energy T enters a transverse uniform magnetic field of induction B. After 3\ 	ext{s}, the kinetic energy of the particle will be

Accepted Answer

T. The magnetic force is always perpendicular to velocity, so it does no work and never changes the kinetic energy of the particle. Hence the kinetic energy remains T.

Question 9

A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched OFF and the same magnetic field is maintained, the electrons move

Accepted Answer

in a circular orbit. With the electric field off, only the magnetic force (perpendicular to velocity) acts on the electrons, causing them to move in a circular orbit.

Question 10

Under the influence of a uniform magnetic field, a charged particle is moving in a circle of radius R with constant speed v. The time period of the motion

Accepted Answer

is independent of both R and v. Since T = \dfrac{2\pi m}{qB}, the time period depends only on the mass, charge and magnetic field, and is independent of both R and v.

Question 11

When a charged particle moving with velocity \vec{v} is subjected to a magnetic field of induction \vec{B}, the force on it is non-zero. This implies that

Accepted Answer

angle between \vec{v} and \vec{B} can have any value other than zero and 180^\circ. Since F = qvB\sin	heta, a non-zero force requires \sin	heta 
eq 0, i.e., 	heta can be any value other than 0^\circ and 180^\circ.

Question 12

An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to

Accepted Answer

\sqrt{\dfrac{v}{B}}. The equivalent current is i = \dfrac{ev}{2\pi r}, so B = \dfrac{\mu_0 i}{2r} = \dfrac{\mu_0 e v}{4\pi r^2}, giving r = \sqrt{\dfrac{\mu_0 e v}{4\pi B}} \propto \sqrt{\dfrac{v}{B}}.

Question 13

A charged particle moves through a magnetic field in a direction perpendicular to it. Then, the

Accepted Answer

speed of the particle remains unchanged. The magnetic force continuously changes the direction of the particle but never does work on it, so the speed (magnitude of velocity) remains unchanged.

Question 14

A charge q moves in a region where electric field \vec{E} and magnetic field \vec{B} both exist, then the force on it is

Accepted Answer

q\vec{E} + q(\vec{v}	imes\vec{B}). The net Lorentz force on the charge is the sum of the electric force and the magnetic force: \vec{F} = q\vec{E} + q(\vec{v}	imes\vec{B}).

Question 15

A charged particle of charge q and mass m enters perpendicularly in a magnetic field B. Kinetic energy of the particle is E, then frequency of rotation is

Accepted Answer

\dfrac{qB}{2\pi m}. Magnetic force provides centripetal force: qvB=\dfrac{mv^2}{r} \Rightarrow \omega=\dfrac{qB}{m}. Since \omega=2\pi
u, the frequency of rotation is 
u=\dfrac{qB}{2\pi m}, independent of the kinetic energy E.

Question 16

A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will

Accepted Answer

move in a circular orbit with its speed unchanged. The magnetic field is perpendicular to the velocity, so the magnetic force F=q\vec{v}	imes\vec{B} always acts perpendicular to \vec{v}, doing no work. The particle moves in a circular orbit with unchanged speed.

Question 17

A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular electric and magnetic fields of strength 20\ 	ext{Vm}^{-1} and 0.5\ 	ext{T} respectively, at right angles to the direction of motion of the electrons. Then, the velocity of electrons must be

Accepted Answer

40\ 	ext{m/s}. For undeviated motion (velocity selector condition), qE=qvB \Rightarrow v=\dfrac{E}{B}=\dfrac{20}{0.5}=40\ 	ext{m/s}.

Question 18

A 10 eV electron is circulating in a plane at right angles to a uniform field of magnetic induction 10^{-4}\ 	ext{Wb/m}^2 (\approx 1.0 gauss). The orbital radius of the electron is

Accepted Answer

11\ 	ext{cm}. \dfrac{1}{2}mv^2=10\ 	ext{eV} \Rightarrow v^2=\dfrac{2	imes 10	imes 1.6	imes 10^{-19}}{9.1	imes 10^{-31}}\approx 3.52	imes 10^{12}, so v\approx 1.88	imes 10^6\ 	ext{m/s}. Then r=\dfrac{mv}{qB}=\dfrac{9.1	imes 10^{-31}	imes 1.88	imes 10^6}{1.6	imes 10^{-19}	imes 10^{-4}}\approx 11\ 	ext{cm}.

Question 19

An electron enters a region where magnetic field (B) and electric field (E) are mutually perpendicular, then

Accepted Answer

it can go undeflected also. The net (Lorentz) force is \vec{F}=q(\vec{E}+\vec{v}	imes\vec{B}). If \vec{E} and \vec{B} are chosen such that the electric and magnetic forces cancel, \vec{F}=0 and the particle passes through undeflected.

Question 20

A charge moving with velocity v in the x-direction is subjected to a field of magnetic induction in the negative x-direction. As a result, the charge will

Accepted Answer

remain unaffected. Force F=qvB\sin	heta; here \vec{v} and \vec{B} are anti-parallel (angle 180^\circ), so \sin 180^\circ=0 and F=0. The charge remains unaffected.

Magnetic Force on Charged Particle in Magnetic Field — NEET Physics PYQ MCQs with Solutions

Questions with solutions