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The magnetic field at the point of intersection of diagonals of a square loop of side L carrying a current I is

Options

  1. Aμ 0 I π L
  2. B2 μ 0 I π L
  3. C2 μ 0 I π L
  4. D2 2 μ 0 I π L

Correct answer

D. 2 2 μ 0 I π L

Step-by-step solution

From Biot-Savart's law, magnetic field due to current carrying conductor is B =   μ 0 4 π i r   sin ⁡ ϕ 1 + sin ⁡ ϕ 2 Where, ϕ 1 = 45 ° , ϕ 2 = 45 ° B = μ 0 4 π   . i L 2 1 2 + 1 2 = μ 0 i 2 2 π L Since there are four sides, the total magnetic field is B net = 4 B = 2 2 μ 0 i π L

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