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

Step 1: Write the formula for work done: 
The work done (\(W\)) in pulling the loop is related to the induced emf and the resistance of the loop: \[ W = F \cdot l = \frac{B^2 v l^2}{R} \] where:  \(B = 40 \, \text{T}\) (magnetic field strength),  \(v = 0.05 \, \text{m/s}\) (velocity of pulling the loop),  \(l = 0.05 \, \text{m}\) (length of one side of the loop, calculated as \(\sqrt{\text{Area}} = \sqrt{25 \, \text{cm}^2}\)), \item \(R = 10 \, \Omega\) (resistance of the loop). 
Step 2: Substitute the values into the formula: 
\[ W = \frac{40^2 \cdot 0.05 \cdot 0.05^2}{10} \] \[ W = \frac{1600 \cdot 0.05 \cdot 0.0025}{10} \] \[ W = \frac{1600 \cdot 0.000125}{10} = \frac{0.2}{10} = 0.001 \, \text{J} \] Step 3: Convert to millijoules: 
\[ W = 1.0 \times 10^{-3} \, \text{J} \]