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PhyHardAdv 2019 • Paper 1

Potential Difference Induced in Moving Parabolic Wire in Magnetic Field

Consider a conducting wire with a parabolic profile, initially given by y=x2y = x^2, moving at a constant velocity V⃗=V0i^\vec{V} = V_0 \hat{i} within a non-uniform magnetic field B⃗=B0(1+(yL)β)k^\vec{B} = B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) \hat{k}, as depicted in the figure. Here, V0V_0, B0B_0, LL, and β\beta represent positive constants. If Δϕ\Delta \phi denotes the potential difference generated across the ends of the wire, select the correct statement(s):

Question Diagram 1

Options

A

∣Δϕ∣=12B0V0L|\Delta \phi| = \frac{1}{2} B_0 V_0 L when β=0\beta = 0

B

∣Δϕ∣=43B0V0L|\Delta \phi| = \frac{4}{3} B_0 V_0 L when β=2\beta = 2

Correct
C

∣Δϕ∣|\Delta \phi| remains unchanged if the parabolic wire is substituted by a straight wire of length 2L\sqrt{2} L initially along y=xy = x

Correct
D

∣Δϕ∣|\Delta \phi| is directly proportional to the length of the wire projected onto the yy-axis.

Topics & Concepts

Electromagnetic InductionFaraday's LawLenz's LawMagnetic Force

Step-by-Step Solution

To determine the correct statement(s), we calculate the induced electromotive force (potential difference) across the ends of the moving wire using the concept of motional EMF.

1. Derivation of the Induced Potential Difference

The velocity of the wire is given by: V⃗=V0i^\vec{V} = V_0 \hat{i}

The non-uniform magnetic field is: B⃗=B0(1+(yL)β)k^\vec{B} = B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) \hat{k}

The motional electric field vector E⃗ind\vec{E}_{\text{ind}} set up along the conductor due to its motion in the magnetic field is: E⃗ind=V⃗×B⃗=(V0i^)×[B0(1+(yL)β)k^]=−V0B0(1+(yL)β)j^\vec{E}_{\text{ind}} = \vec{V} \times \vec{B} = (V_0 \hat{i}) \times \left[ B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) \hat{k} \right] = - V_0 B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) \hat{j}

For an infinitesimal line element along the wire, dl⃗=dxi^+dyj^d\vec{l} = dx \hat{i} + dy \hat{j}, the induced potential difference dϕd\phi across this element is: dϕ=(V⃗×B⃗)⋅dl⃗=−V0B0(1+(yL)β)dyd\phi = (\vec{V} \times \vec{B}) \cdot d\vec{l} = - V_0 B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) dy

The total magnitude of the potential difference developed between the ends of the wire (from y=0y = 0 to y=Ly = L) is: ∣Δϕ∣=∣∫0L−V0B0(1+(yL)β)dy∣=V0B0∫0L(1+yβLβ)dy|\Delta \phi| = \left| \int_0^L - V_0 B_0 \left(1 + \left(\frac{y}{L}\right)^\beta\right) dy \right| = V_0 B_0 \int_0^L \left(1 + \frac{y^\beta}{L^\beta}\right) dy

Evaluating the integral: ∣Δϕ∣=V0B0[y+yβ+1Lβ(β+1)]0L=V0B0(L+Lβ+1)=B0V0L(1+1β+1)|\Delta \phi| = V_0 B_0 \left[ y + \frac{y^{\beta+1}}{L^\beta(\beta+1)} \right]_0^L = V_0 B_0 \left( L + \frac{L}{\beta+1} \right) = B_0 V_0 L \left(1 + \frac{1}{\beta+1}\right)

2. Evaluation of the Options

  • Option A: For β=0\beta = 0: ∣Δϕ∣=B0V0L(1+10+1)=2B0V0L|\Delta \phi| = B_0 V_0 L \left(1 + \frac{1}{0+1}\right) = 2 B_0 V_0 L Hence, Option A is incorrect.

  • Option B: For β=2\beta = 2: ∣Δϕ∣=B0V0L(1+12+1)=43B0V0L|\Delta \phi| = B_0 V_0 L \left(1 + \frac{1}{2+1}\right) = \frac{4}{3} B_0 V_0 L Hence, Option B is correct.

  • Option C: The differential potential difference dϕd\phi depends solely on the yy-component of the wire element (dydy) and the yy-coordinate. Replacing the parabolic wire with a straight wire y=xy = x of length 2L\sqrt{2}L extends from x=0x = 0 to x=Lx = L, which corresponds to y=0y = 0 to y=Ly = L. Because the initial and final yy-limits remain 00 and LL, the integration result for ∣Δϕ∣|\Delta \phi| remains identical. Hence, Option C is correct.

  • Option D: Because of the dependence on yβy^\beta inside the integrand, the expression for ∣Δϕ∣|\Delta \phi| is not directly proportional to the projected length on the yy-axis for a general value of β>0\beta > 0. Hence, Option D is incorrect.

Conclusion

The correct options are B and C.

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