An iron rod of susceptibility \(599\) is subjected to a magnetising field of \(1200~\text{A m}^{-1}\).  The permeability of the material of the rod is:\(\left(\mu_0 = 4 \pi\times 10^{-7}~\text{T m A}^{-1}\right)\)
1. \(8.0\times 10^{-5}~\text{T m A}^{-1}\)
2. \(2.4\pi\times 10^{-5}~\text{T m A}^{-1}\)
3. \(2.4\pi\times 10^{-7}~\text{T m A}^{-1}\)
4. \(2.4\pi\times 10^{-4}~\text{T m A}^{-1}\)

At point A on the earth's surface, the angle of dip is, $\delta =+25°$. At a point B on the earth's surface, the angle of dip is, $\delta =-25°$. We can interpret that:

The relation amongst the three elements of Earth's magnetic field, namely horizontal component H, vertical component V and dip angle$$ $\delta$ is: ($B_E$=total magnetic field):

1. V=$B_E$tan$\delta$, H=$B_E$

2. V=$B_E$sin$\delta$, H=$B_E$cos$\delta$

3. V=$B_E$cos$\delta$, H=$B_E$sin $\delta$

4. V=$B_E$, H=$B_E$tan$\delta$ 

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 250-turn rectangular coil with a length of 2.1 cm and a width of 1.25 cm carries 85 \(\mu\)A and is subjected to a magnetic field with a strength of 0.85 T. What is the work done to rotate the coil by 180 degrees against the torque?

1. 9.1 $\mu J$

2. 4.55 $\mu J$

3. 2.3 $\mu$$J$

4. 1.5 $\mu J$

If ${\theta }_1 and {\theta }_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\theta$ is given by 

(1) $co{t}^2\theta =co{t}^2{\theta }_1+co{t}^2{\theta }_2$

(2) ${\tan }^2\theta ={\tan }^2{\theta }_1+ {\tan }^2{\theta }_2$

(3) $co{t}^2\theta =co{t}^2{\theta }_1-{\tan }^2{\theta }_2$

(4) ${\tan }^2\theta ={\tan }^2{\theta }_1-{\tan }^2{\theta }_2$