JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Electromagnetic Induction

A conducting wire of parabolic shape, initially $y=x^2$, is moving with velocity $\vec{V}=V_0\hat{i}$ in a nonuniform magnetic field $\overrightarrow{\mathrm{B}}={\mathrm{B}}_0\left(1+{\left(\frac{\mathrm{y}}{\mathrm{L}}\right)}^{\beta }\right)\hat{\mathrm{k}}$, as shown in figure. If ${\mathrm{V}}_0,{\mathrm{B}}_0,\mathrm{L}$ and $\beta$ are positive constants and $\Delta \varphi$ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:

1. (A)$|\Delta \varphi |$ remains the same if the parabolic wire is replaced by a straight wire, $\mathrm{y}=\mathrm{x}$ initially, of length $\sqrt{2}\mathrm{L}$✓ Correct
2. (B)$|\Delta \varphi |$ is proportional to the length of the wire projected on the y-axis.✓ Correct
3. (C)$|\Delta \varphi |=\frac{1}{2}{\mathrm{B}}_0{\mathrm{V}}_0\mathrm{L}$ for $\beta =0$
4. (D)$|\Delta \varphi |=\frac{4}{3}{\mathrm{B}}_0{\mathrm{V}}_0\mathrm{L}$ for $\beta =2$✓ Correct

Answer: (A), (B), (D)

Solution

The motional emf is:

$$\int d\varphi ={\int }_0^L{V}_0{B}_0\left(1+\frac{y^{\beta }}{L^{\beta }}\right)dy$$

$$\Delta \varphi =V_0{B}_0\left[L+\frac{L^{\beta +1}}{(\beta +1)L^{\beta }}\right]=V_0{B}_0\left[L+\frac{L}{\beta +1}\right]$$

$$|\Delta \varphi |=B_0{V}_0\left(1+\frac{1}{\beta +1}\right)L$$

Since $|\Delta \varphi |\propto L$ (the y-projection), option (B) is correct.

For $\beta =0$:

$$\Delta \varphi =V_0{B}_0[L+L]=2V_0{B}_{0}L$$

Option (C) is incorrect (states $\frac{1}{2}{B}_0{V}_{0}L$).

For $\beta =2$:

$$\Delta \varphi =V_0{B}_0\left[L+\frac{L}{3}\right]=\frac{4}{3}{V}_0{B}_{0}L$$

Option (D) is correct.

If the wire is replaced by a straight wire $y=x$ of length $\sqrt{2}L$, the y-projection remains L, so $|\Delta \varphi |$ remains the same. Option (A) is correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options