Question

Two parallel wires \(AL\) and \(BM\), placed at a distance \(l\), are connected by a resistor \(R\) and placed in a magnetic field \(B\) perpendicular to the plane of the wires. Another wire \(CD\) connects the two wires perpendicularly and is made to slide with velocity \(v\). Neglect the resistance of all the wires. What is the work done per second needed to slide the wire \(CD\)?

• A. \[ \frac{Blv}{R} \]
• B. \[ \frac{B^2l^2v^2}{R^2} \]
• C. \[ \frac{Bl^2v}{R} \]
• D. \[ \frac{B^2l^2v^2}{R} \]

Hint:

For a sliding rod: \[ \varepsilon=Blv \] \[ I=\frac{Blv}{R} \] and the required mechanical power is \[ P=I^2R=\frac{B^2l^2v^2}{R}. \]

Answer 1

The Correct Option is D

Step 1: Why work is needed.
Sliding the wire $CD$ changes the area of the loop, which changes the flux and drives a current. The external agent must supply power to keep it moving steadily.

Step 2: Energy shortcut.
When the rod moves at constant speed, all the work done per second by the agent ends up dissipated as heat in the resistor. So work done per second equals the power lost in $R$.

Step 3: Motional emf.
The rod of length $l$ moving at speed $v$ across field $B$ sets up \[ \varepsilon=Blv. \]

Step 4: Power dissipated in the resistor.
Using $P=\dfrac{\varepsilon^2}{R}$, \[ P=\frac{(Blv)^2}{R}. \]

Step 5: Simplify.
\[ P=\frac{B^2 l^2 v^2}{R}. \]

Step 6: Read off the answer.
Work done per second to slide the wire equals this power.
\[ \boxed{\dfrac{B^2 l^2 v^2}{R}} \]