ve force (emf), the current in the circuit, the power dissipated in the resistor, and finally, the rate of work done to maintain the motion of the wire.

First, let's calculate the induced emf. According to Faraday's law of electromagnetic induction, the emf $$\varepsilon$$ induced in a wire moving in a magnetic field is given by:

$$\varepsilon = Blv$$

Where:

$$B$$ = Magnetic field strength (in Tesla, T)

$$l$$ = Length of the wire (in meters, m)

$$v$$ = Velocity of the wire (in meters per second, m/s)

Given:

$$B = 0.5\, \text{T}$$

$$l = 1\, \text{m}$$

$$v = 2\, \text{m/s}$$

Let's substitute these values into the formula:

$$\varepsilon = 0.5 \times 1 \times 2 = 1\, \text{V}$$

Next, we calculate the current $$I$$ using Ohm's law, where the resistance $$R$$ is given as 6$$\Omega$$:

$$I = \frac{\varepsilon}{R}$$

Substituting the known values:

$$I = \frac{1\, \text{V}}{6\, \Omega} = \frac{1}{6}\, \text{A}$$

Now, the power dissipated in the resistor, which is also the rate at which work is done to keep the wire moving, can be calculated using the formula for power:

$$P = I^2 R$$

Substituting the values we have:

$$P = \left(\frac{1}{6}\right)^2 \times 6$$

$$P = \frac{1}{36} \times 6 = \frac{1}{6}\, \text{W}$$

Therefore, the rate at which work is being done to keep the wire moving at that speed is $$\frac{1}{6}$$ W.

The correct answer is:

Option B: $$\frac{1}{6}$$ W

Electromagnetic Induction

Continue this paper

Similar Electromagnetic Induction PYQs

More practice from this question