Question

A square loop of area \(25\ cm^2\) has a resistance of \(10\ Ω\). The loop is placed in uniform magnetic field of magnitude \(40.0 \ T\). The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in \(1.0\) sec, will be

• A. $1.0×10^{-3} J $
• B. $ 2.5×10^{−3} J $
• C. $ 5×10^{−3} J $
• D. $ 1.0×10^{−4} J $

Answer 1

The Correct Option is A

To determine the work done in pulling the square loop out of the magnetic field, we can follow these steps:

1. Given:
   • Area of the square loop, A = 25\ cm^2 = 25 \times 10^{-4}\ m^2
   • Resistance of the loop, R = 10\ \Omega
   • Magnetic field strength, B = 40.0\ T
   • Time taken to pull the loop out, t = 1.0\ s
2. The initial magnetic flux, \Phi_i, through the loop when it is fully inside the magnetic field is calculated as:
   \Phi_i = B \cdot A = 40.0\ T \times 25 \times 10^{-4}\ m^2 = 1 \ Wb
3. The final magnetic flux, \Phi_f, when the loop is completely out of the field is:
   \Phi_f = 0 \ Wb
4. The change in magnetic flux, \Delta \Phi, is:
   \Delta \Phi = \Phi_f - \Phi_i = 0 - 1 = -1\ Wb
5. According to Faraday’s Law of Electromagnetic Induction, the induced emf (electromotive force), \epsilon, in the loop is:
   |\epsilon| = \left| \frac{\Delta \Phi}{\Delta t} \right| = \left| \frac{-1}{1\ s} \right| = 1 \ V
6. The induced current, I, in the loop can be calculated using Ohm’s Law:
   I = \frac{\epsilon}{R} = \frac{1\ V}{10\ \Omega} = 0.1\ A
7. The power, P, dissipated in the loop due to the resistance is given by:
   P = I^2 \cdot R = (0.1\ A)^2 \times 10\ \Omega = 0.1\ W
8. The work done, W, in pulling the loop out of the field over time t is:
   W = P \cdot t = 0.1\ W \times 1\ s = 0.1\ J = 1.0 \times 10^{-3} J
9. Therefore, the work done in pulling the loop out of the magnetic field is 1.0 \times 10^{-3}\ J.