A) \[{{10}^{2}}\]
B) 1
C) \[{{10}^{4}}\]
D) \[{{10}^{3}}\]
Correct Answer: C
Solution :
(c) \[{{10}^{4}}\] Here, \[H=2\times {{10}^{3}}A{{m}^{-1}}\], \[B=8\pi \,T,\text{ }{{\mu }_{0}}=4\pi \times {{10}^{-7}}\] Since \[{{\mu }_{r}}=\frac{\mu }{{{\mu }_{0}}}=\frac{\mu H}{{{\mu }_{0}}H}=\frac{B}{{{\mu }_{0}}H}\] \[=\frac{8\pi }{4\pi \times {{10}^{-7}}\times 2\times {{10}^{3}}}={{10}^{4}}\]
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