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A copper rod is moved in a magnetic field. The charge developed across its ends will be proportional to
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Verified Answer
The correct answer is:
rate of change of magnetic flux
When a copper rod is moved in magnetic field, an emf is induced in such a way that current induced will try to oppose the motion of the rod.
As we know,
$$
\begin{array}{lll}
& e \propto \frac{d \phi}{d t} & \\
\Rightarrow & i R \propto \frac{d \phi}{d t} & (\because e=i R) \\
\Rightarrow & \frac{d Q}{d t} \times R \propto \frac{d \phi}{d t} & \left(\because i=\frac{d Q}{d t}\right) \\
\Rightarrow & \frac{d Q}{d t} \propto \frac{d \phi}{d t} &
\end{array}
$$
Hence, charge developed across the ends of the rod will be proportional to rate of change of magnetic flux.
As we know,
$$
\begin{array}{lll}
& e \propto \frac{d \phi}{d t} & \\
\Rightarrow & i R \propto \frac{d \phi}{d t} & (\because e=i R) \\
\Rightarrow & \frac{d Q}{d t} \times R \propto \frac{d \phi}{d t} & \left(\because i=\frac{d Q}{d t}\right) \\
\Rightarrow & \frac{d Q}{d t} \propto \frac{d \phi}{d t} &
\end{array}
$$
Hence, charge developed across the ends of the rod will be proportional to rate of change of magnetic flux.
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