A) \[0.12\times {{10}^{-4}}\,T\]
B) \[0.21\times {{10}^{-4}}T\]
C) \[0.34\times {{10}^{-4}}T\]
D) \[0.87\times {{10}^{-7}}T\]
Correct Answer: C
Solution :
[c] \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{({{r}^{2}}+{{\ell }^{2}})}^{3/2}}}={{B}_{H}}\] Given that \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}\,T\,m{{A}^{-1}}\] \[M=1.34\,A{{m}^{2}}\] \[r=15\,cm=0.15\,m\] and \[\ell =5.0\,cm=0.05\,m\] \[\therefore \,\,\,\,\,{{B}_{H}}={{10}^{-7}}\times \frac{1.34}{{{[{{(0.15)}^{2}}+{{(0.5)}^{2}}]}^{3/2}}}\] \[={{10}^{-7}}\times \frac{1.34}{0.025\sqrt{0.025}}=0.34\times {{10}^{-4}}\,T\]
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