Force Acting on Moving Charges

Practice Questions:

Question 21: moderate

Two very long straight parallel wires carry steady currents i and 2i in opposite directions. The
distance between the wires is d. At a certain instant of time a point charge q is at a point
equidistant from the two wires in the plane of the wires. Its instantaneous velocity \(\overrightarrow{v}\)

is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is :

1. \[\frac{\mu_{0}iqv}{2\pi d}\]

2. \[\frac{\mu_{0}iqv}{\pi d}\]

3. \[\frac{3\mu_{0}iqv}{2\pi d}\]

4. Zero

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Question 22: easy

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 fields, then the electron

1. Speed will decrease

2. Speed will increase

3. Will turn towards right of direction of motion

4. Will turn towards left of direction of motion

View Answer

Since the velocity \(vec{v}\) is parallel to the magnetic field \(vec{B}\), the magnetic force is zero. The electric field exerts a force opposite to the direction of velocity on the negatively charged electron, decreasing its speed.

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Question 23: easy

Which of the following charges has the maximum frequency of revolution in a uniform transverse magnetic field?

1. a proton

2. an alpha particle

3. an electron

4. a neutron

View Answer

Frequency of revolution in a magnetic field is given by \(f = \frac{qB}{2\pi m}\). Since the electron has the highest charge-to-mass ratio \(q/m\) among the charged particles, it has the maximum frequency.

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Question 24: easy

A particle of mass \(M\) and charge \(Q\) moving with velocity \(\vec{v}\) describes a circular path of radius \(R\) when subjected to a uniform transverse magnetic field of induction \(B\). The work done by the field when the particle completes one full circle is:

1. \(BQv2\pi R\)

2. \(\left(\frac{M v^2}{R}\right) 2\pi R\)

3. Zero

4. \(BQ2\pi R\)

View Answer

The magnetic force \(\vec{F} = Q(\vec{v} \times \vec{B})\) is always perpendicular to the velocity \(\vec{v}\). Therefore, power \(P = \vec{F} \cdot \vec{v} = 0\), meaning the work done is always zero.

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Question 25: easy

A magnetic field:

1. Always exerts a force on a charged particle

2. Never exerts a force on a charged particle

3. Exerts a force, if the charged particle is moving across the magnetic field lines

4. Exerts a force, if the charged particle is moving along the magnetic field lines

View Answer

The magnetic force is given by \(F = qvB\sin\theta\). If the particle moves across the field lines, \(\sin\theta \neq 0\), resulting in a non-zero force.

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Question 26: moderate

If a charged particle goes unaccelerated in a region containing electric & magnetic fields:
(a) \(\vec{E}\) must be perpendicular to \(\vec{B}\)
(b) \(\vec{v}\) must be perpendicular to \(\vec{E}\)
(c) \(\vec{v}\) must be perpendicular to \(\vec{B}\)
(d) \(E\) must be equal to \(vB\)

1. (a), (b)

2. (c), (d)

3. (a), (c)

4. (b), (d)

View Answer

For the net force to be zero, \(\vec{F}_e + \vec{F}_b = 0 ⇒ \vec{E} = -(\vec{v} \times \vec{B})\). Since \(\vec{E}\) is the cross product, it must be perpendicular to both \(\vec{B}\) and \(\vec{v}\).

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Question 27: easy

If magnetic field in space is \(1\text{ T } \hat{i}\), electric field is \(10\text{ N/C } \hat{i}\), no gravitational field is present and a charged particle is released from rest from origin, it will:

1. not move at all

2. move in circular path

3. move in a helical path

4. move on a straight line

View Answer

Since the particle starts from rest, its initial magnetic force is zero. The electric field accelerates it along \(\hat{i}\). Because velocity remains parallel to the magnetic field, the magnetic force remains zero, and it continues on a straight line.

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Question 28: easy

Statement-1: In an isolated conductor, free electrons keep on moving but no net magnetic force acts on a conductor in a magnetic field.

Statement-2: In a conductor, the average velocity of thermal motion of electrons is zero. Hence no current flows through the conductor.

1. Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.

2. Both Statement-1 and Statement-2 are true but Statement-2 is not correct explanation of Statement-1.

3. Statement-1 is true but Statement-2 is false.

4. Statement-1 and Statement-2 are false.

View Answer

The net magnetic force on a current-carrying conductor is given by \(F = I L B\). Since average velocity of thermal motion is zero, current \(I = 0\), resulting in zero net force.

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Question 29: moderate

In the product \(\vec{F} = q(\vec{v} \times \vec{B}) = q \vec{v} \times (B_x \hat{i} + B_y \hat{j} + B_0 \hat{k})\), 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 \(vec{B}\)?

1. \(6\hat{i} + 6\hat{j} - 8\hat{k}\)

2. \(-8\hat{i} - 8\hat{j} - 6\hat{k}\)

3. \(-6\hat{i} - 6\hat{j} - 8\hat{k}\)

4. \(8\hat{i} + 8\hat{j} - 6\hat{k}\)

View Answer

Using the relation \(\vec{F} = \vec{v} \times \vec{B}\), we compare vector components: \(4\hat{i} - 20\hat{j} + 12\hat{k} = (4 B_0 - 6 B_y)\hat{i} - (2 B_0 - 6 B_x)\hat{j} + (2 B_y - 4 B_x)\hat{k}\). Testing the values in Option C gives \(B_x = -6\, B_y = -6\, B_0 = -8\), which completely satisfies all equations.

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