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**Hint:**Find the magnetic field due to the current-carrying wire. Use this magnetic field equation to find the induced emf between $EF$ . Using Ohm’s law, find the current in the loop and hence find the force required to maintain the constant speed.

**Complete Step by step solution:**

The equation for the magnetic field around the current-carrying wire at a distance $r$ is given by

$\Rightarrow$ $B=\dfrac{{{\mu }_{o}}I}{2\pi r}$

Emf induced by the current-carrying rod $PQ$ on the rod $EF$ is given by

$\Rightarrow$ $emf=\int\limits_{a}^{b}{BVdr}$

Substituting the already written equation for the magnetic field into the above equation, we get

$\Rightarrow$ $emf=\int\limits_{a}^{b}{\dfrac{{{\mu }_{o}}I}{2\pi r}Vdr}$

After integration and substituting the lower and upper limit into the integrated term, we get

$\Rightarrow$ $emf=\dfrac{{{\mu }_{o}}IV}{2\pi }\ln \left( \dfrac{b}{a} \right)$

By Ohm's law,

$I=\dfrac{V}{R}$

Therefore, using OHM’s law, we get the current due to induced emf as

$i=\dfrac{emf}{R}$

By substituting the derived equation for emf in the above equation for current, we get

$\Rightarrow$ $i=\dfrac{{{\mu }_{o}}IV}{2\pi R}\ln \left( \dfrac{b}{a} \right)$

Now, the force on the induced current-carrying rod due to the magnetic field produced due to the current flowing through the rod $PQ$ is given by

$\Rightarrow$ $F=\int\limits_{a}^{b}{Bidr}$

By substituting the equations for the magnetic field and the induced current, we get

$F=\int\limits_{a}^{b}{\dfrac{{{\mu }_{o}}I}{2\pi r}\dfrac{{{\mu }_{o}}IV}{2\pi R}\ln \left( \dfrac{b}{a} \right)dr}$

Integrating the above equation and substituting the upper and lower limits to the integrated equation gives us

$\Rightarrow$ $F=\dfrac{V}{R}{{\left( \dfrac{{{\mu }_{o}}I}{2\pi }\ln \left( \dfrac{b}{a} \right) \right)}^{2}}$

This is the force required to move the rod $EF$ at a constant velocity $V$ .

**Note:**

Force on a current-carrying conductor is given by

$F=BIL\sin \theta $

Here, $\theta $ is the angle subtended between the conductor and the direction of the magnetic field. In the given question, the angle between the conductor and the magnetic field is found to be ${{90}^{\circ }}$ . So, $\sin \theta $ becomes $1$ and that is the reason we don’t see the sine term in the equation.

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