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

1. 3M/$\mathrm{\pi }$
2. 2M/l$\mathrm{\pi }$
3. M/2
4. M

A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in the earth's horizontal magnetic field of 24 $\mu$T. When a horizontal field of 18 $\mu$T is produced opposite to the earth's field by placing a current-carrying wire, the new time period of the magnet will be:

1. 1 s

2. 2 s

3. 3 s

4. 4 s

A magnet is parallel to a uniform magnetic field. If it is rotated by 60°, the work done is 0.8 J. How much work is done in moving it 30° further?

1. $0.8\times {10}^7 \mathrm{ergs}$ 

2.  0.4 J

3. 8 J 

4. 0.8 ergs

The current of 30 A is flowing in a vertical straight wire. If the horizontal component of the earth's magnetic field is 2 x 10^-5 Tesla, then the position of the null point will be:

1. 0.9 m 

2. 0.3 mm

3. 0.3 cm 

4. 0.3 m

A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10^–4 m² carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10^–2 tesla making an angle of 30° with the axis of the solenoid. The torque on the the solenoid will be:
1. 3 × 10^–3 Nm 
2. 1.5 × 10^–3 Nm 
3. 1.5 × 10^–2 Nm 
4. 3 × 10^–2 Nm

A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from:

A frog can be levitated in a magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog behaves as:
1. paramagnetic 
2. diamagnetic
3. ferromagnetic 
4. none of these

A current carrying coil is placed with its axis perpendicular to N-S direction. Let horizontal component of earth's magnetic field be ${\mathrm{H}}_0$ and magnetic field inside the loop be H. If a magnet is suspended inside the loop, it makes angle $\mathrm{\theta }$ with H. Then $\mathrm{\theta \; equal\; to:}$

1. ${\tan }^{-1}\left(\frac{H_0}{H}\right)$

2. ${\tan }^{-1}\left(\frac{H}{H_{\mathit{0}}}\right)$

3. $\cos e{c}^{-1}\left(\frac{H}{H_{\mathit{0}}}\right)$

4. $co{t}^{-1}\left(\frac{H_0}{H}\right)$