Question

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 ______ \( \times 10^{-6} \, \text{J} \).

Answer 1

Correct Answer: 3

The work done in pulling the loop out of the magnetic field can be calculated using the concept of electromagnetic induction. When the loop is pulled out of the magnetic field, an electromotive force (EMF) is induced which results in a current due to the resistance of the loop. The power loss due to this current is equal to the rate at which work is done.

Step 1: Determine the Induced EMF
The change in magnetic flux (\( \Delta \Phi \)) is given by:
\(\Delta \Phi = B \times A\), where \(B = 0.5 \, \text{T}\) and \(A = 3.6 \times 10^{-3} \, \text{m}^2\).
\(\Delta \Phi = 0.5 \times 3.6 \times 10^{-3} = 1.8 \times 10^{-3} \, \text{Tm}^2\).
Since there are 10 turns, total flux change = \(10 \times 1.8 \times 10^{-3} = 1.8 \times 10^{-2} \, \text{Tm}^2\).

Step 2: Calculate Induced Current
The EMF (\( \mathcal{E} \)) induced 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:
\(I = \frac{\mathcal{E}}{R} = \frac{1.8 \times 10^{-2}}{100} = 1.8 \times 10^{-4} \, \text{A}\).

Step 3: Calculate the Work Done
Work done is equal to the 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 calculated work \(3.24 \times 10^{-6} \, \text{J}\) is indeed \(3.24 \approx 3 \, \times 10^{-6} \, \text{J}\) which fits the expected range of 3,3.

Thus, the work done is \(3.24 \times 10^{-6} \, \text{J}\).

Answer 2

The emf induced in the loop is:
\[\mathcal{E} = N B v \ell,\]
where:
\[v = \frac{\ell}{t}.\]
The current induced in the loop is:
\[i = \frac{\mathcal{E}}{R} = \frac{N B \ell / t}{R}.\]
The force acting is:
\[F = N \cdot i \cdot B \cdot \ell = \frac{N^2 B^2 \ell^2}{R t}.\]
The work done is:
\[W = F \cdot \ell = \frac{N^2 B^2 \ell^2}{R t} \cdot \ell = \frac{N^2 B^2 \ell^3}{R t}.\]
Substitute values:
\[W = \frac{(10)^2 (0.5)^2 (3.6 \times 10^{-3})^2}{100 \cdot 1}.\]
\[W = 3.24 \times 10^{-6} \, \text{J}.\]
Final Answer: $3.24 \times 10^{-6} \, \text{J}$.