A) 4 Hz
B) 2.5 Hz
C) 9 Hz
D) 15 Hz
Correct Answer: D
Solution :
\[v=\frac{1}{2\pi }\sqrt{\frac{M{{B}_{H}}}{I}}\] \[v=\frac{1}{2\pi }\sqrt{\frac{M(B+{{B}_{H}})}{I}}\]where B = Magnetic field due to downward conductor. Then, \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2i}{a}\] \[B={{10}^{-7}}\times \frac{2\times 15}{20\times {{10}^{-2}}}\] \[B=15\mu T\] \[\therefore \] \[\frac{v}{v}=\sqrt{\frac{B+{{B}_{H}}}{B}}\] \[\frac{v}{10}=\sqrt{\frac{15+12}{12}}=1.5\] \[v=15\,Hz\]You need to login to perform this action.
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