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The figure shows a wire sliding on two parallel conducting rails placed at a separation $\prime I\prime $ . A magnetic field $B$ exists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire moving at a constant velocity $v$ ?
(A) $\dfrac{{{B^2}{l^2}v}}{R}$
(B) $\dfrac{{2{B^2}{l^2}v}}{R}$
(C) $\dfrac{{{B^2}{l^2}v}}{{2R}}$
(D) None of these

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Last updated date: 13th Jun 2024
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Answer
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Hint: To solve this question, we will use the expression for magnetic Lorentz force. Also, we will need ohm’s law and Faraday's law for induced emf to find the current induced in the circuit as there is a change in the magnetic field, which results in generating induced emf in a closed loop. Then, by substituting the expression for the current in the Lorentz force equation, we can find the force.
Formula used:
$F = Bil$
$\varepsilon = Blv$
$i = \dfrac{\varepsilon }{R}$

Complete step-by-step solution:
The force experienced by the wire while moving on the parallel conducting wire through the magnetic field is given by the equation,
$F = i\left( {\vec B \times \vec l} \right)$
$F = Bil$
where $i$ is the induced current flowing through the loop.
$\vec B$ is the magnetic field.
and $\vec l$ is the length of the wire.
Now, the current developed inside the loop can be evaluated using ohm’s law.
$i = \dfrac{\varepsilon }{R}$
where emf say $\varepsilon $ is the induced emf due to change in magnetic flux as the wire is moving
and $R$ is the resistance of the loop.
Now, the induced emf is given by faraday’s law for metallic wire as,
$\varepsilon = \dfrac{{d\phi }}{{dt}}$
$ \Rightarrow \varepsilon = \dfrac{{d\left( {Blx} \right)}}{{dt}}$
where $x$ is the distance moved by the wire
$ \Rightarrow \varepsilon = Bl\dfrac{{dx}}{{dt}}$
$\therefore \varepsilon = Blv$
where, $v$ is the velocity of the wire.
At this time, substituting this value for emf in ohm’s law, we will get the current induced in the loop.
$i = \dfrac{\varepsilon }{R}$
$ \Rightarrow i = \dfrac{{Blv}}{R}$
Currently, we can compute the force essential to keep the wire moving with a constant velocity by substituting this current in the expression for Lorentz force.
$F = Bil$
Put the value of $i$ in the above equation we get,
$F = B\left( {\dfrac{{Blv}}{R}} \right)l$
$ \Rightarrow F = \dfrac{{{B^2}{l^2}v}}{R}$
Hence, the force required is found to be $\dfrac{{{B^2}{l^2}v}}{R}$ .

So, option (A) is correct.

Note: We must be having a clear memory of formulas and theories of basic laws and equations to solve. Moreover, these types of questions may come as an open circuit as an alternative to closed-loop as this question. Then no current will be flowing over the circuit and force becomes zero.