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JEE Main 2022 (29 Jul, Shift 1) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2022 (29 Jul, Shift 1)

Q69JEE Main 2022 · 29 JUL · SHIFT 1Easy

Biot-Savart Law & Circular Coils

A closely wound circular coil of radius 5 cm produces a magnetic field of 37.68×10−4 T37.68 \times 10^{-4}\text{ T} at its center. The current through the coil is _________A.

[Given, number of turns in the coil is 100 and π=3.14\pi = 3.14]

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. B=μ0NI2rB = \frac{\mu_0 N I}{2r}, so I=2rBμ0NI = \frac{2rB}{\mu_0 N}.
  2. I=2×0.05×37.68×10−44×3.14×10−7×100=3.768×10−41.256×10−4I = \frac{2\times 0.05\times 37.68\times10^{-4}}{4\times 3.14\times10^{-7}\times 100} = \frac{3.768\times10^{-4}}{1.256\times10^{-4}}
  3. So the current is 3 A3\text{ A}.
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