Question 1

The magnetic field generated along the axis of a solenoid is proportional to:

Accepted Answer

number of turns per unit length in it. For a long solenoid, the magnetic field along its axis is given by B = μ₀ n I, where n is the number of turns per unit length. So the field is directly proportional to the turn density, not the length, current squared, or radius.

Question 2

Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment m⃗. Which configuration has highest net magnetic dipole moment?

Accepted Answer

A. The net magnetic dipole moment is the vector sum of the individual dipole moments. The configuration with all dipole moments aligned in the same direction will have the highest net dipole moment. Without specific diagrams, configuration A is assumed to have this alignment.

Question 3

A bar magnet of length 'ℓ' and magnetic dipole moment 'M' is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be

Accepted Answer

(b) 2/π M. When the bar magnet is bent into an arc, its magnetic dipole moment is proportional to the effective length of the arc. The effective length of the arc is reduced to 2/π of the original length, so the new magnetic dipole moment becomes 2/π times the original dipole moment.

Question 4

A bar magnet of magnetic moment M is placed at right angles to a magnetic induction B. If a force F is experienced by each pole of the magnet, the length of the magnet will be

Accepted Answer

(b) MB/F. The force on each pole of the magnet is given by F = mB, where m is the pole strength. The magnetic moment M is related to the pole strength and length of the magnet as M = m × l. Rearranging, l = M/F × B, which simplifies to MB/F.

Question 5

A short bar magnet of magnetic moment 0.4 J T⁻¹ is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is:

Accepted Answer

-0.064 J. The potential energy of a magnetic dipole in a uniform magnetic field is given by U = -MBcosθ. For stable equilibrium, θ = 0°, so cosθ = 1. Substituting M = 0.4 J T⁻¹ and B = 0.16 T, we get U = -(0.4)(0.16)(1) = -0.064 J.

Question 6

A bar magnet having a magnetic moment of 2×10⁴ J T⁻¹ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6×10⁻⁴ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is:

Accepted Answer

0.6 J. The work done in rotating a magnetic dipole in a magnetic field is given by W = MB(1 - cosθ). Substituting M = 2×10⁴ J T⁻¹, B = 6×10⁻⁴ T, and θ = 60°, we get W = (2×10⁴)(6×10⁻⁴)(1 - cos60°) = 0.6 J.

Question 7

A bar magnet, of magnetic moment M⃗ , is placed in a magnetic field of induction B⃗ . The torque exerted on it is:

Accepted Answer

M⃗ × B⃗. The torque on a magnetic dipole in a magnetic field is given by the vector cross product of the magnetic moment (M⃗) and the magnetic field (B⃗), which is M⃗ × B⃗.

Question 8

At a point A on the earth’s surface the angle of dip, δ = +25°. At a point B on the earth’s surface the angle of dip, δ = -25°. We can interpret that:

Accepted Answer

A is located in the northern hemisphere and B is located in the southern hemisphere.. The angle of dip is positive in the northern hemisphere and negative in the southern hemisphere. Since δ = +25° at A, it indicates A is in the northern hemisphere, and δ = -25° at B indicates B is in the southern hemisphere.

Question 9

The work done in turning a magnet of magnetic moment M by an angle of 90° from the meridian, is n times the corresponding work done to turn it through an angle of 60°. The value of n is given by:

Accepted Answer

2. The work done in turning a magnet in a uniform magnetic field is proportional to the difference in the cosine of the angles. For 90°, the work is proportional to (1 - 0), and for 60°, it is proportional to (1 - 0.5). The ratio is 2:1, so n = 2.

Question 10

An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m⁻¹. The permeability of the material of the rod is: (μ₀ = 4π×10⁻⁷ T m A⁻¹)

Accepted Answer

2.4π×10⁻⁴ T m A⁻¹. The permeability of the material is given by μ = μ₀(1 + χ), where χ is the susceptibility. Substituting μ₀ = 4π×10⁻⁷ T m A⁻¹ and χ = 599, we get μ = 4π×10⁻⁷ × (1 + 599) = 2.4π×10⁻⁴ T m A⁻¹.

Question 11

The relations amongst the three elements of earth’s magnetic field, namely horizontal component H, vertical component V and dip δ are, (Bₑ = total magnetic field):

Accepted Answer

V = Bₑ sinδ, H = Bₑ cosδ. The vertical component (V) and horizontal component (H) of Earth's magnetic field are related to the total magnetic field (Bₑ) and the angle of dip (δ) as V = Bₑ sinδ and H = Bₑ cosδ, based on trigonometric resolution of the magnetic field vector.

Question 12

If θ₁ and θ₂ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip θ is given by:

Accepted Answer

tan²θ = tan²θ₁ + tan²θ₂. The true angle of dip is related to the apparent angles of dip in two perpendicular planes by the formula tan²θ = tan²θ₁ + tan²θ₂. This is derived from the vector components of the Earth's magnetic field.

Question 13

The magnetic susceptibility is negative for:

Accepted Answer

diamagnetic material only. Diamagnetic materials have a negative magnetic susceptibility because they create an induced magnetic field in the opposite direction to the applied magnetic field.

Question 14

There are four light-weight-rod samples A, B, C, D separately suspended by threads. A bar magnet is slowly brought near each sample and the following observations are noted:
(i) A is feebly repelled
(ii) B is feebly attracted
(iii) C is strongly attracted
(iv) D remains unaffected
Which on

Accepted Answer

C is of a ferromagnetic material. C is strongly attracted to the magnet, which is a characteristic property of ferromagnetic materials. Ferromagnetic materials exhibit strong attraction to magnetic fields.

Question 15

The magnetic moment of a diamagnetic atom is:

Accepted Answer

equal to zero. Diamagnetic atoms have no unpaired electrons, so their net magnetic moment is zero.

Question 16

Electromagnets are made of soft iron because soft iron has:

Accepted Answer

low retentivity and low coercive force. Soft iron is used in electromagnets because it has low retentivity and low coercive force, allowing it to magnetize and demagnetize quickly, which is essential for efficient operation.

Question 17

If a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is:

Accepted Answer

repelled by both the poles. Diamagnetic substances are repelled by both poles of a magnet because they create an induced magnetic field in the opposite direction to the external magnetic field.

Question 18

Curie temperature is the temperature above which:

Accepted Answer

ferromagnetic material becomes paramagnetic material. The Curie temperature is the temperature above which a ferromagnetic material loses its ferromagnetic properties and becomes paramagnetic due to thermal agitation disrupting the alignment of magnetic domains.

Question 19

Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show:

Accepted Answer

paramagnetism. Above the Curie temperature, ferromagnetic materials like nickel lose their ferromagnetic properties and behave as paramagnetic materials due to the randomization of magnetic domains.

Question 20

A diamagnetic material in a magnetic field moves:

Accepted Answer

from stronger to the weaker parts of the field. Diamagnetic materials are repelled by magnetic fields, so they move from regions of stronger magnetic field to weaker magnetic field.

NEET Physics: Magnetism and Matter questions with solutions

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