JEE Advanced
- Exam
- JEE Advanced
- Subject
- Physics
- Chapter
- Magnetic Properties of Matter
JEE Advanced · Physics · Magnetic Properties of Matter
A magnetic needle lying parallel to the magnetic field requires W...
A magnetic needle lying parallel to the magnetic field requires units of work to turn it through an angle The torque required to maintain the needle in this position will be
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
Work done by magnet to turn from angle
\begin{array}{l}<br>\begin{array}{l}<br>\theta_{1} \text { to } \theta_{2} \\<br>\mathrm{~W}=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right) \\<br>=\mathrm{MB}\left(\cos 0^{{}^{{}^{\circ}}}-\cos 45^{{}^{{}^{\circ}}}\right)<br>\end{array} \\
=\mathrm{MB}\left(1-\frac{1}{\sqrt{2}}\right)=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \mathrm{MB}
\end{array}
Also torque acting on the magnet
\begin{array}{l}<br>\begin{array}{l}<br>\theta_{1} \text { to } \theta_{2} \\<br>\mathrm{~W}=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right) \\<br>=\mathrm{MB}\left(\cos 0^{{}^{{}^{\circ}}}-\cos 45^{{}^{{}^{\circ}}}\right)<br>\end{array} \\
=\mathrm{MB}\left(1-\frac{1}{\sqrt{2}}\right)=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \mathrm{MB}
\end{array}
Also torque acting on the magnet