Q51 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity ($g=10 \mathrm{~m} / \mathrm{s}^{2}$) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire when it travells a vertical distance of 200 m is $\_\_\_\_$ mV.
- Correct answerA$20 \sqrt{10}$
- B$0.2 \sqrt{10}$
- C$200 \sqrt{10}$
- D$2 \sqrt{10}$
Solution
For a conducting wire falling in a magnetic field, induced EMF is: $\mathcal{E} = BLv$
After falling through height $h = 200$ m, velocity: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 200} = \sqrt{4000} = 20\sqrt{10}$ m/s
Converting magnetic field: $B = 0.5$ Gauss $= 0.5 \times 10^{-4}$ T
Induced EMF: $\mathcal{E} = (0.5 \times 10^{-4}) \times 20 \times 20\sqrt{10} = 200 \times 10^{-4}\sqrt{10} = 0.02\sqrt{10}$ V $= 20\sqrt{10}$ mV
After falling through height $h = 200$ m, velocity: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 200} = \sqrt{4000} = 20\sqrt{10}$ m/s
Converting magnetic field: $B = 0.5$ Gauss $= 0.5 \times 10^{-4}$ T
Induced EMF: $\mathcal{E} = (0.5 \times 10^{-4}) \times 20 \times 20\sqrt{10} = 200 \times 10^{-4}\sqrt{10} = 0.02\sqrt{10}$ V $= 20\sqrt{10}$ mV
Hence, the correct answer is option: 'A'.