Magnetism and Matter — NCERT Solutions

For a magnetic dipole in a uniform field:

$$U=-mB\cos\theta$$
$$\tau=mB\sin\theta$$

Here $m=1.5\,\text{J T}^{-1}$ and $B=0.22\,\text{T}$.

(a) Work required to turn the magnet

If the magnet initially lies aligned with the field, then $\theta=0^\circ$.

(i) Turned so that $m$ is normal to the field

Then $\theta=90^\circ$.

Initial energy:
$$U_i=-mB=-1.5\times 0.22=-0.33\,\text{J}$$

Final energy:
$$U_f=0$$

Work done by external torque:
$$W=U_f-U_i=0-(-0.33)=0.33\,\text{J}$$

(ii) Turned opposite to the field

Then $\theta=180^\circ$.

Final energy:
$$U_f=+mB=+0.33\,\text{J}$$

Work done:
$$W=U_f-U_i=0.33-(-0.33)=0.66\,\text{J}$$

(b) Torque on the magnet

(i) When normal to the field

$$\tau=mB\sin 90^\circ=1.5\times 0.22=0.33\,\text{N m}$$

(ii) When opposite to the field

$$\tau=mB\sin 180^\circ=0$$

So the required results are: work $0.33\,\text{J}$ and $0.66\,\text{J}$; torque $0.33\,\text{N m}$ and $0$ respectively.

For the solenoid, the magnetic moment is

$$m=NIA$$

Given:

• $N=2000$
• $I=4.0\,\text{A}$
• $A=1.6\times 10^{-4}\,\text{m}^2$

So,

$$m=2000\times 4.0\times 1.6\times 10^{-4}$$

$$m=8000\times 1.6\times 10^{-4}=1.28\,\text{A m}^2$$

If the solenoid is suspended freely in a horizontal plane, it behaves like a magnetic dipole and will tend to align with an external magnetic field. The field exerts no net force on it in a uniform field, but it can experience a torque that turns it.

For a short bar magnet at distance $r$ far from its centre:

• On the axis,

$$B_A=\frac{\mu_0}{4\pi}\frac{2m}{r^3}$$

• On the equatorial line,

$$B_E=-\frac{\mu_0}{4\pi}\frac{m}{r^3}$$

Given:

• $m=0.48\,\text{J T}^{-1}=0.48\,\text{A m}^2$
• $r=10\,\text{cm}=0.10\,\text{m}$
• $\mu_0/4\pi = 10^{-7}$

(a) On the axis

$$B_A=10^{-7}\times \frac{2\times 0.48}{(0.10)^3}$$

$$B_A=10^{-7}\times \frac{0.96}{10^{-3}}=10^{-7}\times 960=9.6\times 10^{-5}\,\text{T}$$

Direction: along the magnetic moment of the magnet on the axial line.

(b) On the equatorial line

$$B_E=10^{-7}\times \frac{0.48}{(0.10)^3}$$

$$B_E=10^{-7}\times \frac{0.48}{10^{-3}}=10^{-7}\times 480=4.8\times 10^{-5}\,\text{T}$$

Direction: opposite to the magnetic moment on the equatorial line.