Magnetism and Matter - Result Question 7
10. The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the meridian, is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is given by
(a) 2
(b) 1
(c) 0.5
(d) 0.25
[1995]
Topic 2: The Earth’s Magnetism, Magnetic
Materials and their Properties11. An iron rod of susceptibility 599 is subjected to a magnetising field of $1200 A m^{-1}$. The permeability of the material of the rod is :
$(\mu_0=4 \pi \times 10^{-7} T m A^{-1})$
(a) $8.0 \times 10^{-5} T m A^{-1}$
[2020]
(b) $2.4 \pi \times 10^{-5} Tm A^{-1}$
(c) $2.4 \pi \times 10^{-7} T m A^{-1}$
(d) $2.4 \pi \times 10^{-4} T m A^{-1}$
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Answer:
Correct Answer: 10. (a)
Solution:
- (a) Magnetic moment $=M$; Initial angle through which magnet is turned $(\theta_1)=90^{\circ}$ and final angle through which magnet is turned $(\theta_2)=$ $60^{\circ}$. Work done in turning the magnet through $90^{\circ}(W_1)=MB(\cos 0^{\circ}-\cos 90^{\circ})$ $=M B(1-0)=M B$.
Similarly, $W_2=M B(\cos 0^{\circ}-\cos 60^{\circ})$
$=M B(1-\frac{1}{2})=\frac{M B}{2}$
$\therefore W_1=2 W_2$ or $n=2$.
