In a certain region static electric and magnetic fields exist. The magnetic field is given by B → = B 0 i ^ + 2 j ^ - 4 k ^ . If a test charge moving with a velocity v → = v 0 3 i ^ - j ^ + 2 k ^ experiences no force in that region, then the electric field in the region, in SI units, is:

E → = - v 0 B 0 14 j ^ + 7 k ^

E → = - v 0 B 0 i ^ + j ^ + 7 k ^

E → = - υ 0 B 0 3 i ^ - 2 j ^ - 4 k ^

EASY

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A charge $q$ is moving with velocity $v$ produces an electric field which satisfy Coulomb's law or Gauss's law or both or none.

Important Questions on Magnetic Fields due to Electric Current

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

An electron enters a chamber in which a uniform magnetic field is present as shown. Ignore gravity
Question Image

During its motion inside the chamber

EASY

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

An electron is moving with a velocity

2 \times 10^{6} m / s

along positive

x

-direction in the uniform electric field of

8 \times 10^{7} V / m

applied along positive

y

-direction. The magnitude and direction of a uniform magnetic field (in tesla) that will cause the electrons to move undeviated along its original path is

EASY

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is:

EASY

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A proton, a deuteron and an

α

-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A proton and an

α

- particle (with their masses in the ratio of

1 : 4

and charges in the ratio of

1 : 2

) are accelerated from rest through a potential difference

V .

If a uniform magnetic field

(B)

is set up perpendicular to their velocities, the ratio of the radii

r_{p} : r_{α}

of the circular paths described by them will be:

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii

r_{e}, r_{p}, r_{α}

respectively in a uniform magnetic field B. The relation between

r_{e}, r_{p}, r_{α}

is:

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

In a certain region static electric and magnetic fields exist. The magnetic field is given by

\vec{B} = B_{0} (\hat{i} + 2 \hat{j} - 4 \hat{k})

. If a test charge moving with a velocity

\vec{v} = v_{0} (3 \hat{i} - \hat{j} + 2 \hat{k})

experiences no force in that region, then the electric field in the region, in SI units, is:

HARD

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A charged particle is introduced at the origin

x = 0, y = 0, z = 0

with a given initial velocity

\vec{v}

. A uniform electric field

\vec{E}

and a uniform magnetic field

\vec{B}

exist everywhere. The velocity

\vec{v}

, electric field

\vec{E}

and a uniform magnetic field

\vec{B}

are given in the columns below

Column 1	Column 2	Column 3
1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v} = 0$	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(iv) $\vec{E} = E_{0} \hat{x}$	(S) $\vec{B} = B_{0} \hat{z}$

In which case will the particle describe a helical path with axis along positive z direction?

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

An electron is moving in a circular path under the influence of a transverse magnetic field of

3.57 \times 10^{- 2} T

. If, the value of

\frac{e}{m}

1 .76 \times 10^{11} C {kg}^{- 1}

, the frequency of revolution of the electron is

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

An electron enters a chamber in which a uniform magnetic field is present as shown.
Question Image

An electric field of appropriate magnitude is also applied so that the electron travels un-deviated without any change in its speed through the chamber. We are ignoring gravity. Then, the direction of the electric field is,

HARD

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

In the

xy - plane

, the region

y > 0

has a uniform magnetic field

B_{1} \hat{k}

and the region

y < 0

has another uniform magnetic field

B_{2} \hat{k}

. A positively charged particle is projected from the origin along the positive

y - axis

with speed

v_{0} = π m s^{- 1}

t = 0

, as shown in the figure. Neglect gravity in this problem. Let

t = T

be the time when the particle crosses the

x - axis

from below for the first time. If

B_{2} = 4 B_{1}

, the average speed of the particle, in

m s^{- 1}

, along the

x - axis

in the time interval

T

is ________.

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HARD

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

If one were to apply the Bohr model to a particle of mass

' m'

and charge

' q'

moving in a plane under the influence of a magnetic field 'B', the energy of the charged particle in the

n^{t h}

level will be:

HARD

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A charged particle is introduced at the origin

x = 0, y = 0, z = 0

with a given initial velocity

\vec{v}

. A uniform electric field

\vec{E}

and a uniform magnetic field

\vec{B}

exist everywhere. The velocity

\vec{v}

, electric field

\vec{E}

and a uniform magnetic field

\vec{B}

are given in the columns below

Column 1	Column 2	Column 3
1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v} = 0$	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(iv) $\vec{E} = E_{0} \hat{x}$	(S) $\vec{B} = B_{0} \hat{z}$

In which case would the particle move in a straight line along the negative direction of y axis

EASY

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. if the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:

HARD

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A charged particle is introduced at the origin

x = 0, y = 0, z = 0

with a given initial velocity

\vec{v}

. A uniform electric field

\vec{E}

and a uniform magnetic field

\vec{B}

exist everywhere. The velocity

\vec{v}

, electric field

\vec{E}

and a uniform magnetic field

\vec{B}

are given in the columns below

Column 1	Column 2	Column 3
1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v} = 0$	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(iv) $\vec{E} = E_{0} \hat{x}$	(S) $\vec{B} = B_{0} \hat{z}$

In which case will the particle move in a straight line with a constant velocity?

EASY

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

Ionized hydrogen atoms and

α

-particles with same momenta enters perpendicular to a constant magnetic field

B

. The ratio of their radii of their paths

r_{H} : r_{α}

will be

EASY

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

α

-particle consists of

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A particle of charge

Q

moves with a velocity

v = a \hat{i}

in a magnetic field

B = b \hat{j} + c \hat{k}

where

a, b

and

c

are constants. The magnitude of the force experienced by the particle is

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A rectangular region of dimensions $w \times I (w < < I)$ has a constant magnetic field into the plane of the paper as shown. On one side the region is bounded by a screen. On the other side positive ions of mass $m$ and charge $q$ are accelerated from rest and towards the screen by a parallel plate capacitor at constant potential difference $V < 0$ , and come out through a small hole in the upper plate. Which one of the following statements is correct regarding the charge on the ions that hit the screen?

Question Image

MEDIUM

Physics>Electricity and Magnetism>Magnetic Fields due to Electric Current>Magnetic Force

A proton (mass $m$ ) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$ . The field occupies a region of space by width $d$ . If $α$ be the angle of deviation of proton from the initial direction of motion (see figure), the value of $\sin α$ will be:
Question Image

A charge q is moving with velocity v produces an electric field which satisfy Coulomb's law or Gauss's law or both or none.

Important Questions on Magnetic Fields due to Electric Current

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A charge $q$ is moving with velocity $v$ produces an electric field which satisfy Coulomb's law or Gauss's law or both or none.