Magnetism Question 44
Question: A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through $180 ^\circ$ to deflect the magnet by $30 ^\circ$ from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through $270^\circ$ to deflect the magnet $30^\circ$ from magnetic meridian. The ratio of magnetic moments of magnets is
Options:
A) 1 : 5
B) 1 : 8
C) 5 : 8
D) 8 : 5
Show Answer
Answer:
Correct Answer: C
Solution:
Let M1 and M2 be the magnetic moments of magnets and H the horizontal component of earth’s field. We have $ \tau =MH\sin \theta $ . If f is the twist of wire, then $ \tau =C\varphi , $ C being restoring couple per unit twist of wire
$ \Rightarrow \ C\varphi =MH\sin \theta $ Here $ {\varphi _{1}}=(180^{o}-30^{o})=150^{o} $
$ =150\times \frac{\pi }{180} $ rad $ {\varphi _{2}}=(270^{o}-30^{o})=240^{o} $
$ =240\times \frac{\pi }{180} $ rad So, $ C{\varphi _{1}}=M _{1}H\sin \theta $ (For deflection $ \theta =30^{o} $ of I magnet) $ C{\varphi _{2}}=M _{2}H\sin \theta $ (For deflection $ \theta =30^{o} $ of II magnet) Dividing $ \frac{{\varphi _{1}}}{{\varphi _{2}}}=\frac{M _{1}}{M _{2}} $
$ \Rightarrow \frac{M _{1}}{M _{2}}=\frac{{\varphi _{1}}}{{\varphi _{2}}}=\frac{150\times ( \frac{\pi }{180} )}{240\times ( \frac{\pi }{180} )}=\frac{15}{24}=\frac{5}{8} $
$ \Rightarrow M _{1}:M _{2}=5:8 $
BETA
