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JEE Main 2019 (10 Apr, Shift 2) · Magnetic Field of Straight Wires

Magnetic Field of Straight Wires question from JEE Main 2019 (10 Apr, Shift 2)

Q60JEE Main 2019 · 10 APR · SHIFT 2Medium

Magnetic Field of Straight Wires

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is:

[Take μ\mu0 = 4π\pi × 10–7 NA–2]

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Answer: (b)

  1. Equilateral triangle of side aa: B=9μ0I2πaB = \frac{9\mu_0 I}{2\pi a}.
  2. B=9×4π×10−7×102π×1=18×10−6B = \frac{9 \times 4\pi \times 10^{-7} \times 10}{2\pi \times 1} = 18 \times 10^{-6} T
  3. So the answer is (b) 18 μ18\ \muT.
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