The horizontal component of the earth's magne | vedclass

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(D) The rod is placed in the north-south direction and moves towards the east. The velocity vector $\vec{v}$ is directed towards the east.The vertical

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$10$

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(D) The rod is placed in the north-south direction and moves towards the east. The velocity vector $\vec{v}$ is directed towards the east.The vertical component of the earth's magnetic field is $B_V = B_H 	an \phi$,where $B_H = 3 	imes 10^{-4} \, T$ and $\phi = 53^\circ$.Thus,$B_V = (3 	imes 10^{-4}) 	imes (4/3) = 4 	imes 10^{-4} \, T$.The induced emf $e$ in a rod moving through a magnetic field is given by $e = B_V v l$,where $v = 10 \, cm/s = 0.1 \, m/s$ and $l = 0.25 \, m$.Substituting the values: $e = (4 	imes 10^{-4} \, T) 	imes (0.1 \, m/s) 	imes (0.25 \, m) = 10^{-5} \, V$.Since $1 \, V = 10^6 \, \mu V$,we have $e = 10^{-5} 	imes 10^6 \, \mu V = 10 \, \mu V$.

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