A conducting wire of parabolic shape, initially y = x2, is moving with velocity v = v_0\widehat i in a non-uniform magnetic field B = B_0( 1 + ( y \over L )^\beta )\widehat k , as shown in figure. If V0, B0, L and \beta are positive constants and \Delta \phi is the potential difference developed between the ends of the wire, then the correct statement(s) is/are

| \Delta \phi | = 4 \over 3B_0V_0L for \beta = 2

| \Delta \phi | remains the same if the parabolic wire is replaced by a straight wire, y =x initially, of length \sqrt 2 L

| \Delta \phi | = 1 \over 2B_0V_0L for \beta = 0

| \Delta \phi | is proportional to the length of the wire projected on the y-axis.

Electromagnetic Induction PYQ — JEE Advanced 2019 · Shift 1 · Q42 with Solution

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Correct answer: A, B, D

Motional potential difference across a moving wire

For a conductor moving with velocity $\vec v$ in a magnetic field $\vec B$ , the potential difference between its ends is

\Delta \phi = \int (\vec v \times \vec B)\cdot d\vec l

in magnitude/sign depending on the chosen end points.

Here,

\vec v = v_0\hat i, \qquad \vec B = B_0\left(1+\left(\frac{y}{L}\right)^\beta\right)\hat k

So,

\vec v \times \vec B = v_0\hat i \times B_0\left(1+\left(\frac{y}{L}\right)^\beta\right)\hat k

Since $\hat i \times \hat k = -\hat j$ ,

\vec v \times \vec B = -v_0B_0\left(1+\left(\frac{y}{L}\right)^\beta\right)\hat j

Thus the integrand depends only on $dy$ :

(\vec v\times \vec B)\cdot d\vec l = -v_0B_0\left(1+\left(\frac{y}{L}\right)^\beta\right)dy

Hence,

\Delta \phi = -v_0B_0\int_{y_1}^{y_2}\left(1+\left(\frac{y}{L}\right)^\beta\right)dy

So the magnitude depends only on the end-point $y$ -coordinates, not on the detailed shape of the wire.

Endpoints of the given parabola

The wire is initially of parabolic shape $y=x^2$ . From the figure/implied standard endpoint choice, the ends are at

(0,0) \quad \text{and} \quad (L,L)

because $y=x^2$ passes through $(0,0)$ and $(L,L)$ when $L=1$ would not be general; more properly the intended segment is from $x=0$ to $x=1$ scaled? But the options clearly indicate the relevant $y$ -range is $0$ to $L$ , and option B uses a straight wire $y=x$ of length $\sqrt2 L$ , i.e. joining $(0,0)$ to $(L,L)$ .

Therefore we take

y_1=0,\qquad y_2=L

Thus

|\Delta \phi| = v_0B_0\int_0^L \left(1+\left(\frac{y}{L}\right)^\beta\right)dy

Compute:

|\Delta \phi| = v_0B_0\left[\int_0^L dy + \int_0^L \left(\frac{y}{L}\right)^\beta dy\right]

= v_0B_0\left[L + \frac{1}{L^\beta}\int_0^L y^\beta dy\right]

= v_0B_0\left[L + \frac{1}{L^\beta}\cdot \frac{L^{\beta+1}}{\beta+1}\right]

= v_0B_0L\left(1+\frac{1}{\beta+1}\right)

= v_0B_0L\cdot \frac{\beta+2}{\beta+1}

This is the general result.

Check option A

For $\beta=2$ ,

|\Delta \phi|=v_0B_0L\cdot \frac{2+2}{2+1}=v_0B_0L\cdot \frac43

So,

|\Delta \phi|=\frac43 B_0v_0L

Therefore A is correct.

Check option B

If the wire is replaced by a straight wire $y=x$ of length $\sqrt2 L$ , then its endpoints are still $(0,0)$ and $(L,L)$ .

Since

\Delta \phi = \int (\vec v\times\vec B)\cdot d\vec l

and the integrand reduces to a function of $y$ times $dy$ , the result depends only on the endpoints in $y$ :

|\Delta \phi| = v_0B_0\int_0^L\left(1+\left(\frac yL\right)^\beta\right)dy

So replacing the parabolic wire by any other wire joining the same two endpoints gives the same potential difference.

Therefore B is correct.

Check option C

For $\beta=0$ ,

B=B_0(1+1)=2B_0

Hence

|\Delta \phi| = v_0B_0L\cdot \frac{0+2}{0+1}=2B_0v_0L

Alternatively directly,

|\Delta \phi| = v_0\int_0^L 2B_0\,dy = 2B_0v_0L

But option C says

|\Delta \phi|=\frac12 B_0v_0L

which is wrong.

Therefore C is incorrect.

Check option D

From

\Delta \phi = -v_0B_0\int_{y_1}^{y_2}\left(1+\left(\frac{y}{L}\right)^\beta\right)dy

we see the emf depends on the integration over $y$ , i.e. on the end-point separation along the $y$ -direction. In particular, for a given field form, it is determined by the projection of the wire between the endpoints on the $y$ -axis, not by the actual curved length.

Thus among the given choices, the intended statement is that it is proportional to the wire length projected on the $y$ -axis (for the present geometry/endpoints).

Therefore D is correct.

Final answer

The correct options are

\boxed{A,\ B,\ D}

More from Electromagnetic Induction

View written solutionFree

Correct answer: A, B, D

Motional potential difference across a moving wire

For a conductor moving with velocity $\vec v$ in a magnetic field $\vec B$ , the potential difference between its ends is

\Delta \phi = \int (\vec v \times \vec B)\cdot d\vec l

in magnitude/sign depending on the chosen end points.

Here,

\vec v = v_0\hat i, \qquad \vec B = B_0\left(1+\left(\frac{y}{L}\right)^\beta\right)\hat k

So,

\vec v \times \vec B = v_0\hat i \times B_0\left(1+\left(\frac{y}{L}\right)^\beta\right)\hat k

Since $\hat i \times \hat k = -\hat j$ ,

\vec v \times \vec B = -v_0B_0\left(1+\left(\frac{y}{L}\right)^\beta\right)\hat j

Thus the integrand depends only on $dy$ :

(\vec v\times \vec B)\cdot d\vec l = -v_0B_0\left(1+\left(\frac{y}{L}\right)^\beta\right)dy

Hence,

\Delta \phi = -v_0B_0\int_{y_1}^{y_2}\left(1+\left(\frac{y}{L}\right)^\beta\right)dy

So the magnitude depends only on the end-point $y$ -coordinates, not on the detailed shape of the wire.

Endpoints of the given parabola

The wire is initially of parabolic shape $y=x^2$ . From the figure/implied standard endpoint choice, the ends are at

(0,0) \quad \text{and} \quad (L,L)

Therefore we take

y_1=0,\qquad y_2=L

Thus

|\Delta \phi| = v_0B_0\int_0^L \left(1+\left(\frac{y}{L}\right)^\beta\right)dy

Compute:

|\Delta \phi| = v_0B_0\left[\int_0^L dy + \int_0^L \left(\frac{y}{L}\right)^\beta dy\right]

= v_0B_0\left[L + \frac{1}{L^\beta}\int_0^L y^\beta dy\right]

= v_0B_0\left[L + \frac{1}{L^\beta}\cdot \frac{L^{\beta+1}}{\beta+1}\right]

= v_0B_0L\left(1+\frac{1}{\beta+1}\right)

= v_0B_0L\cdot \frac{\beta+2}{\beta+1}

This is the general result.

Check option A

For $\beta=2$ ,

|\Delta \phi|=v_0B_0L\cdot \frac{2+2}{2+1}=v_0B_0L\cdot \frac43

So,

|\Delta \phi|=\frac43 B_0v_0L

Therefore A is correct.

Check option B

If the wire is replaced by a straight wire $y=x$ of length $\sqrt2 L$ , then its endpoints are still $(0,0)$ and $(L,L)$ .

Since

\Delta \phi = \int (\vec v\times\vec B)\cdot d\vec l

and the integrand reduces to a function of $y$ times $dy$ , the result depends only on the endpoints in $y$ :

|\Delta \phi| = v_0B_0\int_0^L\left(1+\left(\frac yL\right)^\beta\right)dy

So replacing the parabolic wire by any other wire joining the same two endpoints gives the same potential difference.

Therefore B is correct.

Check option C

For $\beta=0$ ,

B=B_0(1+1)=2B_0

Hence

|\Delta \phi| = v_0B_0L\cdot \frac{0+2}{0+1}=2B_0v_0L

Alternatively directly,

|\Delta \phi| = v_0\int_0^L 2B_0\,dy = 2B_0v_0L

But option C says

|\Delta \phi|=\frac12 B_0v_0L

which is wrong.

Therefore C is incorrect.

Check option D

From

\Delta \phi = -v_0B_0\int_{y_1}^{y_2}\left(1+\left(\frac{y}{L}\right)^\beta\right)dy

Thus among the given choices, the intended statement is that it is proportional to the wire length projected on the $y$ -axis (for the present geometry/endpoints).

Therefore D is correct.

Final answer

The correct options are

\boxed{A,\ B,\ D}

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