Magnetic Effects of Current - NEET Physics Questions
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Magnetic Effects of Current

Practice Questions:
Question 101: 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 102: easy

\(\varepsilon_0\) and \(\mu_0\) are the electric permittivity and magnetic permeability of free space respectively. If the corresponding quantities of a medium are \(2\varepsilon_0\) and \(1.5\mu_0\) respectively, the refractive index of the medium will nearly be

1. 3
2. 2
3. \(\sqrt{2}\)
4. \(\sqrt{3}\)
View Answer

The speed of light in free space is \(c = \frac{1}{\sqrt{\varepsilon_0\mu_0}}\) and in a medium is \(v = \frac{1}{\sqrt{\varepsilon\mu}}\). The refractive index \(n = \frac{c}{v} = \sqrt{\frac{\varepsilon\mu}{\varepsilon_0\mu_0}} = \sqrt{2 \times 1.5} = \sqrt{3}\).

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Question 103: easy

A long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be (\(g = 9.8\text{ ms}^{-2}\))

1. 2.75 A
2. 1.75 A
3. 2.45 A
4. 2.25 A
View Answer

For the wire to be suspended, the upward magnetic force must equal gravity: \(I L B = mg ⇒ I = \frac{mg}{LB}\). Plugging in the values: \(I = \frac{0.250 \times 9.8}{2 \times 0.7} = 1.75\text{ A}\).

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Question 104: easy

For the hollow thin cylindrical current carrying straight pipe which statement is correct:

1. Magnetic field inside the pipe is not zero.
2. Magnetic field outside the pipe is zero.
3. Electric field outside the pipe is zero.
4. Electric field on the surface of pipe is zero.
View Answer

Inside the hollow pipe, the magnetic field is zero according to Ampere's law. Since a current-carrying pipe is electrically neutral, the electric field outside the pipe is zero.

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Question 105: moderate

A helium nucleus is moving in a circular path of radius \(0.8\text{ m}\). If it takes \(2\text{ sec}\) to complete one revolution, the magnetic field produced at the centre of the circle is:

1. \(\mu_0 \times 10^{-19}\text{ T}\)
2. \(\frac{10^{-19}}{\mu_0}\text{ T}\)
3. \(2 \times 10^{-19}\text{ T}\)
4. \(\frac{2 \times 10^{-19}}{\mu_0}\text{ T}\)
View Answer

Current is \(I = \frac{q}{T} = \frac{2e}{2} = e = 1.6 \times 10^{-19}\text{ A}\). The magnetic field at the centre is \(B = \frac{\mu_0 I}{2R} = \frac{\mu_0 (1.6 \times 10^{-19})}{2(0.8)} = \mu_0 \times 10^{-19}\text{ T}\).

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Question 106: 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 107: easy

A thin circular wire carrying a current \(I\) has a magnetic moment \(M\). The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment of:

1. \(M\)
2. \(\frac{4}{\pi^2} M\)
3. \(\frac{4}{\pi} M\)
4. \(\frac{\pi}{4} M\)
View Answer

For circle: \(M = I \pi R^2\) where \(2\pi R = L ⇒ R = \frac{L}{2\pi}\), so \(M = \frac{I L^2}{4\pi}\). For square of side \(a = \frac{L}{4}\): \(M' = I a^2 = \frac{I L^2}{16}\). Thus, \(M' = \frac{\pi}{4} M\).

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Question 108: 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 109: 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 110: 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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