Question 134: Magnetism and Matter

Question

A short bar magnet placed with its axis at $30^{\circ}$ with an external field of 800 Gauss, experiences a torque of $0.016 \mathrm{~N} . \mathrm{m}$. The work done in moving it from most stable to most unstable position is $\alpha 	imes 10^{-3} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ .

Accepted Answer

When a magnetic dipole like a bar magnet is placed in a magnetic field (B) at an angle ( $	heta$ ), it experiences a torque.
$$ 	au=\mathrm{M} \mathrm{~B} \sin 	heta $$
Where, $\mathrm{M}=$ magnetic moment of the magnet, $\mathrm{B}=$ magnetic field strength, $	heta=$ Angle between the magnet's axis and the field.
The magnetic field is given in Gauss. We must convert it to the standard SI unit, Tesla (T).
1 Tesla $=10000$ Gauss $\left(10^4 \mathrm{G}ight)$
So, $\mathrm{B}=800$ Gauss $=800 	imes 10^{-4} \mathrm{~T}=0.08 \mathrm{~T}$
Using $	au=\mathrm{MB} \sin 	heta$ :
$$ 0.016=M 	imes(0.08) 	imes \sin \left(30^{\circ}ight) $$
$$\Rightarrow $$ $0.016=\mathrm{M} 	imes 0.08 	imes 0.5$
$$\Rightarrow $$ $\mathrm{M}=\frac{0.016}{0.04}$
$$\Rightarrow $$ $$ \mathrm{M}=0.4 \mathrm{~A} \cdot \mathrm{~m}^2 $$
The potential energy of the magnetic dipole is given as $\mathrm{U}=-\mathrm{MB} \cos 	heta$
The magnet is aligned with the magnetic field, i.e., $	heta_1=0^{\circ}$. The potential energy is minimum.
The magnet is aligned against (opposite to) the magnetic field, i.e., $	heta_2=180^{\circ}$. The potential energy is maximum.
The work done to rotate a magnet from an initial angle $	heta_1$ to a final angle $	heta_2$ is equal to the change in its potential energy.
$$ \mathrm{W}=\mathrm{MB}\left(\cos 	heta_1-\cos 	heta_2ight) $$
1. $	heta_1=0^{\circ}$
2. $	heta_2=180^{\circ}$
So, the work done is,
$$ \mathrm{W}=(0.4) 	imes(0.08) 	imes\left[\cos \left(0^{\circ}ight)-\cos \left(180^{\circ}ight)ight] $$
$$\Rightarrow $$ $\mathrm{W}=0.032 	imes[1-(-1)]=0.032 	imes 2$
$$\Rightarrow $$ $\mathrm{W}=0.064 \mathrm{~J}=64 	imes 10^{-3} \mathrm{~J}=\alpha 	imes 10^{-3} \mathrm{~J}$
$$\Rightarrow $$ $\alpha=64$

Magnetism and Matter

A short bar magnet placed with its axis at 30^{\circ} with an external field of 800 Gauss, experiences a torque of 0.016 \mathrm{~N} . \mathrm{m}. The work done in moving it from most stable to most unstable position is \alpha \times 10^{-3} \mathrm{~J}. The value of \alpha is \_\_\_\_ .

Integer / Numeric Answer Type