A square loop PQRS having 10 turns, area \(3.6 \times 10^{-3} \, \text{m}^2\), and resistance \(100 \, \Omega\) is slowly and uniformly being pulled out of a uniform magnetic field of magnitude \(B = 0.5 \, \text{T}\) as shown. Work done in pulling the loop out of the field in \(1.0 \, \text{s}\) is ______ \( \times 10^{-6} \, \text{J} \).
The energy expended in withdrawing the loop from the magnetic field is determined by the principle of electromagnetic induction. As the loop is extracted, a voltage (electromotive force, EMF) is induced. This EMF drives a current through the loop, owing to its resistance. The energy dissipated by this current equates to the rate of work performed.
Step 1: Determine Induced EMF The magnetic flux change (\( \Delta \Phi \)) is calculated as: \(\Delta \Phi = B \times A\), with \(B = 0.5 \, \text{T}\) and \(A = 3.6 \times 10^{-3} \, \text{m}^2\). \(\Delta \Phi = 0.5 \times 3.6 \times 10^{-3} = 1.8 \times 10^{-3} \, \text{Tm}^2\). For 10 turns, the total flux change is \(10 \times 1.8 \times 10^{-3} = 1.8 \times 10^{-2} \, \text{Tm}^2\).
Step 2: Calculate Induced Current The induced EMF (\( \mathcal{E} \)) is given by: \(\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t} = -10 \times \frac{1.8 \times 10^{-3}}{1} = -1.8 \times 10^{-2} \, \text{V}\). The induced current \(I\) is calculated as: \(I = \frac{\mathcal{E}}{R} = \frac{1.8 \times 10^{-2}}{100} = 1.8 \times 10^{-4} \, \text{A}\).
Step 3: Calculate Work Done Work done equals energy dissipated: \(W = I^2 R \Delta t\). \(W = (1.8 \times 10^{-4})^2 \times 100 \times 1 = 3.24 \times 10^{-6} \, \text{J}\).
Verification The computed work \(3.24 \times 10^{-6} \, \text{J}\) is consistent with the expected range, approximating \(3 \times 10^{-6} \, \text{J}\).
Therefore, the work performed is \(3.24 \times 10^{-6} \, \text{J}\).
