Q74 / 90 · JEE Main 2022 (28 Jun Shift 2) · Physics

A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be (Assume the coil to be short circuited.)

  • A
    Doubled
  • Correct answer
    B
    The same
  • C
    Quadrupled
  • D
    Halved

Solution

\(\mathrm{P}=\frac{\varepsilon^2}{R}\) where, \(\varepsilon=\) inducedemf \(=-\frac{d \phi}{d t}\) \(\phi=\mathrm{NBA}\) \(\varepsilon=\mathrm{NA} \frac{\mathrm{dB}}{\mathrm{dt}}\) also, \(R \propto \frac{1}{\mathrm{r}^2}\) where, \(\mathrm{R}=\) resistance and \(\mathrm{r}=\) radius of wire \(\begin{aligned} & \mathrm{P} \propto \frac{\mathrm{N}^2 \mathrm{r}^2}{\mathrm{H}^2} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=1 \\ & \mathrm{P}_1=\mathrm{P}_2 \end{aligned}\) This means if the number of turns were quadrupled and the wire radius halved, the electrical power dissipated would be remains same.

Hence, the correct answer is option: 'B'.

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