Q67 / 75 · JEE Main 2026 (23 January Shift 2) · Physics
A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed perpendicular to plane of loop. The loop is converted into a square loop in 0.5 s. The EMF induced in the loop is $\_\_\_\_$ mV.
- A13.2
- Correct answerB1.32
- C8.25
- D6.6
Solution
Circular loop with radius $r = 0.07$ m in magnetic field $B = 0.2$ T.
Initial flux through circular loop: $\Phi_1 = B \times \pi r^2 = 0.2 \times \pi(0.07)^2 = 0.000980\pi$ Wb
The loop is converted to a square maintaining the same perimeter: $C = 2\pi r = 0.14\pi$ m
Side of square: $a = \frac{0.14\pi}{4} = 0.035\pi$ m
Final flux through square loop: $\Phi_2 = B \times a^2 = 0.2 \times (0.035\pi)^2 = 0.2 \times 0.001225\pi^2$ Wb
Evaluating: $\Phi_1 = 0.2 \times 0.0154 = 0.00308$ Wb
$\Phi_2 = 0.2 \times 0.001225 \times 9.87 = 0.00242$ Wb
Change in flux: $\Delta\Phi = 0.00308 - 0.00242 = 0.00066$ Wb
Average EMF: $\varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{0.00066}{0.5} = 0.00132$ V $= 1.32$ mV
Initial flux through circular loop: $\Phi_1 = B \times \pi r^2 = 0.2 \times \pi(0.07)^2 = 0.000980\pi$ Wb
The loop is converted to a square maintaining the same perimeter: $C = 2\pi r = 0.14\pi$ m
Side of square: $a = \frac{0.14\pi}{4} = 0.035\pi$ m
Final flux through square loop: $\Phi_2 = B \times a^2 = 0.2 \times (0.035\pi)^2 = 0.2 \times 0.001225\pi^2$ Wb
Evaluating: $\Phi_1 = 0.2 \times 0.0154 = 0.00308$ Wb
$\Phi_2 = 0.2 \times 0.001225 \times 9.87 = 0.00242$ Wb
Change in flux: $\Delta\Phi = 0.00308 - 0.00242 = 0.00066$ Wb
Average EMF: $\varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{0.00066}{0.5} = 0.00132$ V $= 1.32$ mV
Hence, the correct answer is option: 'B'.