A copper rod is moved in a magnetic field. The charge developed across its ends will be proportional to

A copper rod is moved in a magnetic field. The charge developed across its ends will be proportional to
  1. magnetic flux
  2. rate of change of magnetic flux
  3. $1 /$ velocity of the rod
  4. $1 /$ magnitude of the magnetic field

Solution

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

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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