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.You need to login to perform this action.
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