Ans (a).

Magnetic moment, M = 1.5 J T^-1 ,

Magnetic field strength, B = 0.22 T

(i) Initial angle between the axis and the magnetic field, 0₁ = 0°

Final angle between the axis and the magnetic field, 0₂ = 90°

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

W = -MB(cos0₂ – cos 0₁)

= -1.5 x 0.22(cos 90° – cos 0°)

= -0.33(0-1)

= 0.33 J

(ii) Initial angle between the axis and the magnetic field, 0₁= 0°

Final angle between the axis and the magnetic field, 0₂ = 180°

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

W = -MB(cos0₂ – cos 0₁)

= -1.5 x 0.22(cos 180 – cos 0°)

= —0.33(—1 -1)

= 0.66 J

Ans (b).

For case (i):      0 = 0₂= 90⁰

Therefore ,Torque,       τ =  MBsin 6

= 1.5×0.22 sin 90°

= 0.33 J

For case (ii): 0 = 0₂= 180°

Therefore Torque,         τ = MB sin 0

= MBsin 180° = 0 J

Number of turns in the solenoid, n = 800,

Area of cross-section, A = 2.5 x 10^-4 m²

Current in the solenoid, I = 3.0 A

A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.

The magnetic moment associated with the given current-carrying solenoid is calculated as:

M = n I A = 800 x 3 x 2.5 x 10-⁴ = 0.6 J T^–1

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, T = 4.5 x 10^-2 J

Angle between the bar magnet and the external magnetic field, 0 = 30° Torque is related to magnetic moment (M) as: T = MB sin 0

Therefore , M = T/Bsin 0

= (4.5 x 10^-2 )/(0.25 x sin 30º)  =0.36 J T⁻¹

Hence, the magnetic moment of the magnet is 0.36 IT^-1.

Ans (a).

The three independent quantities conventionally used for specifying earth’s magnetic field are:

1. Magnetic declination,
2. Angle of dip, and
3. Horizontal component of earth’s magnetic field.

Ans (b).

The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.

Ans (c).

It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole.

Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.

Ans (d).

If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.

Ans (e).
Magnetic moment, M = 8 x 10²² J T⁻¹

Radius of earth, r = 6.4 x 10⁶ m

Magnetic field strength , B = (μ0 /4π) x ( M/r³)

Where, μ0 = Permeability of free space = 4π x 10⁻⁷ T m A^-1

Therefore , B= (4π x 10⁻⁷ x 8 x 10²²)/ (4π x 6.4 x 10⁶) =0.3 G

This quantity is of the order of magnitude of the observed field on earth.

Ans (f).

Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.

Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full scale deflection, I_g = 3 mA = 3 x 10⁻³ A Range of the voltmeter is 0, which needs to be converted to 18 V.

Therefore ,V = 18 V

Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:

R = V/I_g – G

= [18 /(3 x 10⁻³) ]-12 = 6000 -12 = 5988 Ω

Hence, a resistor of resistance 5988 Ω is to be connected in series with the galvanometer.