Induced EMF in Equilateral Triangular Loop Entering Magnetic Field

To determine the variation of the induced electromotive force (emf), $E$, as a function of the position $x$ of the loop, we analyze the magnetic flux $\Phi(x)$ linked with the conducting loop as it enters and leaves the triangular magnetic field region.

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1. Geometry and Coordinate Setup

• Magnetic Field Region: An equilateral triangle of height $L$ located in $x \in [0, L]$.

  • The vertex of the field region is at $x = 0$.
  • The base of the field region is at $x = L$.
  • The width of the magnetic field region at any coordinate $x' \in [0, L]$ is: $w_B(x') = 2 x' \tan 30^\circ = \frac{2x'}{\sqrt{3}}$

• Conducting Loop (PQR): An equilateral triangle of height $L$ moving in the $+x$-direction with velocity $v$.

  • At $t = 0$, vertex P is at $x = 0$.
  • At position $x = vt$, the loop occupies the spatial interval $x' \in [x-L, x]$.
  • The width of the loop at any coordinate $x'$ is: $w_{\text{loop}}(x') = 2(x - x') \tan 30^\circ = \frac{2(x - x')}{\sqrt{3}}$

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2. Derivation of Overlap Area $A(x)$ and Induced EMF $E(x)$

By Faraday's law of electromagnetic induction, the induced emf is given by: $E(x) = -\frac{d\Phi}{dt} = -B v \frac{dA}{dx}$

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Phase 1: $0 \le x \le L$ (Loop entering the field)

The field region and the loop overlap in $x' \in [0, x]$. The overlapping width at $x'$ is $w(x') = \min\left(w_B(x'), w_{\text{loop}}(x')\right)$.

Equating $w_B(x') = w_{\text{loop}}(x')$ yields: $\frac{2x'}{\sqrt{3}} = \frac{2(x - x')}{\sqrt{3}} \implies x' = \frac{x}{2}$

Thus, the area of overlap $A(x)$ is: $A(x) = \int_0^{x/2} \frac{2x'}{\sqrt{3}} \, dx' + \int_{x/2}^x \frac{2(x - x')}{\sqrt{3}} \, dx' = \frac{x^2}{4\sqrt{3}} + \frac{x^2}{4\sqrt{3}} = \frac{x^2}{2\sqrt{3}}$

Taking the derivative with respect to $x$: $\frac{dA}{dx} = \frac{x}{\sqrt{3}}$

Therefore, the induced emf in this interval is: $E(x) = -\frac{B v}{\sqrt{3}} x$

• At $x = 0$: $E(0) = 0$
• At $x = L$: $E(L) = -\frac{B v L}{\sqrt{3}}$

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Phase 2: $L \le x \le 2L$ (Loop leaving the field)

The field region ($x' \in [0, L]$) and the loop ($x' \in [x-L, x]$) overlap in $x' \in [x-L, L]$. Since $x \in [L, 2L]$, the point $x' = \frac{x}{2}$ lies within the interval $[x-L, L]$.

The total area of overlap $A(x)$ is: $A(x) = \int_{x-L}^{x/2} \frac{2x'}{\sqrt{3}} \, dx' + \int_{x/2}^L \frac{2(x - x')}{\sqrt{3}} \, dx'$

Evaluating the integrals: $A(x) = \frac{1}{\sqrt{3}} \left[ \left(\frac{x^2}{4} - (x - L)^2\right) + \left(2xL - L^2 - \frac{3x^2}{4}\right) \right]$ $A(x) = \frac{1}{\sqrt{3}} \left( -\frac{3x^2}{2} + 4xL - 2L^2 \right)$

Taking the derivative with respect to $x$: $\frac{dA}{dx} = \frac{1}{\sqrt{3}} (-3x + 4L)$

Therefore, the induced emf in this interval is: $E(x) = -B v \frac{dA}{dx} = \frac{B v}{\sqrt{3}} (3x - 4L)$

• At $x = L$: $E(L) = -\frac{B v L}{\sqrt{3}}$
• At $x = \frac{4}{3}L$: $E\left(\frac{4}{3}L\right) = 0$
• At $x = 2L$: $E(2L) = \frac{2 B v L}{\sqrt{3}}$

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3. Conclusion and Comparison with Options

1. For $0 \le x \le L$, $E(x)$ decreases linearly from $0$ to $-\frac{B v L}{\sqrt{3}}$.
2. For $L \le x \le 2L$, $E(x)$ increases linearly, crossing zero at $x = \frac{4}{3}L$, and reaches a positive maximum at $x = 2L$ with magnitude twice that at $x = L$.

This behavior is correctly depicted in Option (A).