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Alternating Current

Loop A of rad...

Question

Loop $A$ of radius $r (<< R)$ moves towards loop $B$ with a constant velocity $v$ in such a way that their planes are always parallel. What is the distance between the two loops $(x)$ when the induced e.m.f in loop $A$ is maximum?

R

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\frac{R}{\sqrt{2}}

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\frac{R}{2}

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R (1 - \frac{1}{\sqrt{2}})

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Solution

The correct option is C $\frac{R}{2}$
Let $i$ be the current flowing in the loop $B$ .

Magnetic field at loop $A$ due to loop $B$ is given by

$B_{A} = \frac{μ_{0} i R^{2}}{2 (R^{2} + x^{2})^{3 / 2}}$

So, magnetic flux passing through loop $A$ is

$ϕ_{A} = B_{A} (π r^{2})$

$ϕ_{A} = \frac{μ_{0} i R^{2}}{2 (R^{2} + x^{2})^{3 / 2}} (π r^{2})$

$ϕ_{A} = \frac{μ_{0} π i r^{2} R^{2}}{2 (R^{2} + x^{2})^{3 / 2}}$

Applying Faraday's law of induction,

$E_{A} = - \frac{d ϕ_{A}}{d t}$

$E_{A} = - \frac{μ_{0} π i r^{2} R^{2}}{2} (\frac{- 3}{2}) (R^{2} + x^{2})^{- 5 / 2} (2 x)$

$E_{A} = \frac{3 μ_{0} π i r^{2} R^{2}}{2} \frac{x}{(R^{2} + x^{2})^{5 / 2}}$

As we know that $E_{A}$ will be maximum when, $\frac{d E_{A}}{d x} = 0$

$∴ \frac{d E_{A}}{d x} = \frac{d}{d x} \frac{x}{(R^{2} + x^{2})^{5 / 2}} = 0$

$[1 \times (R^{2} + x^{2})^{5 / 2}] - [\frac{5 x}{2} (R^{2} + x^{2})^{3 / 2} (2 x)] = 0$

$R^{2} + x^{2} - 5 x^{2} = 0$

$x = \pm \frac{R}{2}$

Hence, option $(C)$ is correct.

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Alternating Current

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Loop A of radius r (<<R) moves towards loop B with a constant velocity v in such a way that their planes are always parallel. What is the distance between the two loops (x) when the induced e.m.f in loop A is maximum?

Loop $A$ of radius $r (<< R)$ moves towards loop $B$ with a constant velocity $v$ in such a way that their planes are always parallel. What is the distance between the two loops $(x)$ when the induced e.m.f in loop $A$ is maximum?