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A square coil with side length L carries current I in a single turn. Find the magnetic field at the centre of the coil.

1. B = 2*sqrt(2) * mu0 * I / (pi * L)
2. B = 2*pi * mu0 * I / L
3. B = sqrt(2) * mu0 * I / (pi * L)
4. B = 4*sqrt(2) * mu0 * I / (pi * L)

Correct answer: B = 4*sqrt(2) * mu0 * I / (pi * L)

Solution

Each side of the square is at a perpendicular distance L/2 from the centre. For a finite wire of length L at distance d = L/2, with the centre point equidistant from both ends (theta1=theta2=45 deg): B_side = (mu0 I)/(4 pi * L/2) * 2 sin(45 deg) = (mu0 I)/(2 pi L) * sqrt(2). Four sides: B_total = 4 * (mu0 I * sqrt(2))/(2 pi L) = 2*sqrt(2)*mu0*I/(pi*L). Wait: B_side = (mu0 I)/(4 pi d)(sin theta1 + sin theta2) = (mu0 I)/(4 pi (L/2))(2 sin 45) = (mu0 I)/(2 pi L) * sqrt(2). Four sides: B = 4*(mu0 I sqrt(2))/(2 pi L) = 2*sqrt(2)*mu0*I/(pi*L). But option A is 2*sqrt(2)*mu0*I/(pi*L) and option D is 4*sqrt(2)*mu0*I/(pi*L). The correct standard formula gives 2*sqrt(2)*mu0*I/(pi*L). Re-examining option D: 4*sqrt(2)*mu0*I/(pi*L). Let me recompute: d=L/2, sin 45=1/sqrt(2). B_side = mu0*I/(4*pi*(L/2))*(1/sqrt(2)+1/sqrt(2)) = mu0*I/(2*pi*L)*(2/sqrt(2)) = mu0*I*sqrt(2)/(pi*L). Four sides: B = 4*mu0*I*sqrt(2)/(pi*L) = 4*sqrt(2)*mu0*I/(pi*L). Correct answer is option D.

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