Magnetism and Matter - Result Question 7

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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

======= ####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

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/magnetism-and-matter/magnetism-and-matter—result-question-7.md (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:

  1. (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 $n=2$.