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

Concept: A rod of length \(l\) moving with speed \(v\) in a magnetic field \(B\) develops a motional emf \[ \varepsilon=Blv. \] Current in the circuit: \[ I=\frac{\varepsilon}{R}. \] Mechanical power supplied equals electrical power dissipated.

Step 1: Find the induced emf. \[ \varepsilon=Blv. \]

Step 2: Calculate the current in the circuit. \[ I = \frac{\varepsilon}{R} = \frac{Blv}{R}. \]

Step 3: Find the electrical power dissipated. \[ P=I^2R. \] Substituting \(I\), \[ P = \left( \frac{Blv}{R} \right)^2R \] \[ = \frac{B^2l^2v^2}{R}. \] Since the rod moves with constant velocity, the external agent must supply the same power. \[\begin{aligned} \boxed{ \frac{B^2l^2v^2}{R} } \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.