Magnetism Question 42
Question: An iron rod of volume $ {{10}^{-4}}m^{3} $ and relative permeability 1000 is placed inside a long solenoid wound with 5 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod is
Options:
A) $ 10,,A\cdot m^{2} $
B) $ 15,,Am^{2} $
C) $ 20,,A\cdot m^{2} $
D) $ 25,,Am^{2} $
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Answer:
Correct Answer: D
Solution:
We have, $ B={\mu _{0}}(H+I) $ or $ I=\frac{B-{\mu _{0}}H}{{\mu _{0}}} $ or $ I=\frac{(\mu -{\mu _{0}})H}{{\mu _{0}}}=( \frac{\mu }{{\mu _{0}}}-1 )H $
$ I=({\mu _{r}}-1)H $ For a solenoid of n-turns per unit length and current i H = ni
$ \
Therefore \ I=({\mu _{r}}-1)ni $
$ =(1000-1)\times 500\times 0.5 $
$ I=2.5\times 10^{5}A{{m}^{-1}} $
$ \
Therefore \ \ $ Magnetic moment M=IA $ M=2.5\times 10^{5}\times {{10}^{-4}} $ = 25 Am2
BETA
