A conducting square frame of side a and a long straight wire carrying current I | TestHub

A conducting square frame of side $a$ and a long straight wire carrying current $I$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity $V$ . The e.m.f induced in the frame (when the centre of the frame is at a distance $x$ from the wire) will be proportional to :

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Answer:

Solution:

The potential difference across AB is

$V_{A} - V_{B} = B_{1} a . V .$

$\Rightarrow \frac{μ_{i}}{2 π (x - \frac{a}{2})} a V$

The potential difference across CD is

$V_{C} - V_{D} = B_{2} a . V$

$B_{2} = \frac{μ_{0} i}{2 π (x + \frac{a}{2})}$

$V_{C} - V_{D} = \frac{μ_{0} i}{2 π (x + \frac{a}{2})} a V$

Net Potential difference $= \frac{μ_{i} a V}{2 π} (\frac{1}{x - \frac{a}{2}} - \frac{1}{x + \frac{a}{2}})$

$(V_{A} - V_{B}) - (V_{C} - V_{D}) = \frac{μ i a}{2 π} (\frac{2 a}{x^{2} - \frac{a^{2}}{4}})$

$\propto \frac{1}{4 x^{2} - a^{2}} \propto \frac{1}{(2 x + a) (2 x - a)}$

Physics - EMI/AC Question with Solution | TestHub

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