A) 245
B) 250
C) 252
D) 255
Correct Answer: D
Solution :
(d) 255 Here, \[H=1500A{{m}^{-1}},\phi =2.4\times {{10}^{-5}}\] weber \[A=0.5\text{ }c{{m}^{2}}=0.5\times {{10}^{-4\text{ }}}{{m}^{2}}\] \[\therefore B=\frac{\phi }{A}=\frac{2.4\times {{10}^{-5}}}{0.5\times {{10}^{-4}}}=4.8\times {{10}^{-1}}T\] And \[\mu =\frac{B}{H}=\frac{4.8\times {{10}^{-1}}}{1500}=3.2\times {{10}^{-4}}\] So relative permeability \[{{\mu }_{r}}=\frac{\mu }{{{\mu }_{0}}}=\frac{3.2\times {{10}^{-4}}}{4\pi \times {{10}^{-7}}}\] \[=0.255\times {{10}^{3}}=255\]
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