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

Practice Questions:
Question 121: easy

A circular coil of radius \(R\) having current \(I\) is placed in a uniform magnetic field \(B\). If the angle between the area vector of the coil and the magnetic field is \(60^circ\), then the torque on the coil will be:

1. \[\frac{\pi R^2 I B}{2}\]
2. \[\frac{\sqrt{3}\pi R^2 I B}{2}\]
3. \(\pi R^2 I B\)
4. Zero
View Answer

The torque is given by \(\tau = MBsin\theta\), where \(M = I A = I(\pi R^2)\) and \(\theta = 60^\circ\). Thus, \(tau = I(\pi R^2)Bsin 60^\circ = \frac{\sqrt{3}\pi R^2 I B}{2}\).

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Question 122: easy

If a proton has velocity \((2\hat{i} + 3\hat{k})\) m/s and it is subjected to a magnetic field of \(4\hat{i}\) T, then its:

1. Speed will not change
2. Path will not change
3. Velocity will remain same
4. Momentum will remain same
View Answer

Since the magnetic force \(\vec{F} = q(\vec{v} \times \vec{B})\) is always perpendicular to the velocity, the work done is zero. Hence, the kinetic energy and speed remain constant.

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Question 123: easy

A bar magnet has length \(3\text{ cm}\), cross-sectional area \(2\text{ cm}^2\) and magnetic moment \(3\text{ A m}^2\). The intensity of magnetisation of bar magnet is

1. \(2 \times 10^5\text{ A/m}\)
2. \(3 \times 10^5\text{ A/m}\)
3. \(4 \times 10^5\text{ A/m}\)
4. \(5 \times 10^5\text{ A/m}\)
View Answer

Intensity of magnetisation \(I = \frac{M}{V}\). Here, volume \(V = A \times L = (2 \times 10^{-4}\text{ m}^2) \times (3 \times 10^{-2}\text{ m}) = 6 \times 10^{-6}\text{ m}^3\). Thus, \(I = \frac{3}{6 \times 10^{-6}} = 5 \times 10^5\text{ A/m}\).

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Question 124: easy

A charge \( q = 1.6 \times 10^{-12} \text{ C} \) moving with speed of \( v \text{ m s}^{-1} \) crosses electric field \( |\vec{E}| = 6 \times 10^4 \text{ V m}^{-1} \) and magnetic field \( |\vec{B}| = 1.2 \text{ T} \). The electric field and magnetic fields are crossed and velocity \( v \) is also perpendicular to both. If the charge particle crosses both fields undeflected, the value of \( v \) is

1. \( 7.2 \times 10^5 \)
2. \( 7.2 \times 10^4 \)
3. \( 5 \times 10^5 \)
4. \( 5 \times 10^4 \)
View Answer

For a particle to cross perpendicular electric and magnetic fields undeflected, the net force must be zero, which requires \( qE = qvB ⇒ v = \frac{E}{B} \). Substituting the given values: \( v = \frac{6 \times 10^4}{1.2} = 5 \times 10^4 \text{ m/s} \).

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Question 125: easy

Consider the following statements:

A. Magnetic force on a moving charged particle is always non-zero.

B. Magnetic force can change kinetic energy of a charged particle.

C. Magnetic force can change linear momentum of a charged particle.

D. A charged particle at rest does not feel magnetic force on it.

The correct statement(s) is/are

1. Only C
2. A, B and C
3. C and D
4. B, C and D
View Answer

Magnetic force is perpendicular to velocity, so it does no work and KE remains constant. It can change the direction of velocity (and thus momentum). For a particle at rest (\(v = 0\)), magnetic force is zero.

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Question 126: easy

A long solenoid having number of turns per unit length 200 carries a current of \(2.5 \text{ A}\), the magnetic field at the end of the solenoid is

1. \(6.28 \times 10^{-4} \text{ T}\)
2. \(3.14 \times 10^{-4} \text{ T}\)
3. \(6.28 \times 10^{-5} \text{ T}\)
4. \(3.14 \times 10^{-5} \text{ T}\)
View Answer

The magnetic field at the end of a long solenoid is \(B_{\text{end}} = \frac{1}{2} \mu_0 n I\). Substituting the given values: \(B_{\text{end}} = \frac{1}{2} (4\pi \times 10^{-7}) (200) (2.5) = 3.14 \times 10^{-4} \text{ T}\).

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Question 127: easy

Assertion (A): A magnetic field can accelerate a charge particle.

Reason (R): A steady current carrying wire does not generate an electric field outside it.

In the light of the above statements, choose the correct answer from the options given below.

1. Both (A) and (R) are true and (R) is the correct explanation of (A)
2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

A magnetic field exerts a force perpendicular to velocity, changing its direction, hence accelerating the particle. Outside a steady current-carrying wire, there is no net charge, so the electric field is zero. Both statements are true, but they are unrelated.

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Question 128: easy

The substance in which net magnetic dipole moment of an atom is zero, is

1. Paramagnetic
2. Diamagnetic
3. Ferromagnetic
4. Para or ferromagnetic depending on temperature
View Answer

In diamagnetic substances, the magnetic moments of individual electrons cancel each other out, resulting in a net magnetic dipole moment of zero for each atom.

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Question 129: easy

In crossed electric and magnetic field, the velocity of charged particle which passes undeflected through the region may be (where \(E\) is electric field and \(B\) is magnetic field)

1. \(v = E^2 B\)
2. \(v = \frac{E}{B}\)
3. \(v = \frac{E^2}{B}\)
4. \(v = \frac{B}{E}\)
View Answer

For a charged particle to pass undeflected in crossed fields, the electric force must balance the magnetic force: \(qE = qvB ⇒ v = \frac{E}{B}\).

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Question 130: easy

A charged particle enters a magnetic field at right angles to the magnetic field. The field exists for a length equal to 1.5 times the radius of circular path of the circle. The particle will be deviated from its path by angle

1. 90°
2. \(sin^{-1} \left( \frac{2}{3} \right)\)
3. 30°
4. 180°
View Answer

Since the width of the magnetic field \(d = 1.5R > R\), the particle cannot cross the field to the other side. It will complete a semi-circular path inside the field and emerge from the same side it entered, yielding a deviation of \(180^\circ\).

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