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Q.Obtain an expression for torque acting on a rectangular current loop, when placed inclined at an angle 'theta' with the direction of magnetic field 'B'.

Jammu Kashmir JkboseJKBOSE Class 12 Annual Regular Examination 2023Subjective· 3mImportance★★★★★
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Resolving the forces on the two current-carrying sides of a rectangular loop placed in a uniform field B, the net torque about the loop's axis works out to tau = NIABsin(theta), where theta is the angle between the loop's normal (magnetic moment) and B.

Consider a rectangular loop PQRS of sides a (length) and b (breadth), carrying current I, placed in a uniform magnetic field B, such that the normal to the loop (n_hat) makes an angle theta with B.

The forces on the two sides of length b, which lie along the axis of rotation, are equal, opposite, and collinear, so they produce no torque. The two sides of length a experience forces F = IaB, equal and opposite, but NOT collinear - they act along lines separated by a perpendicular distance of b*sin(theta) (since the loop is tilted at angle theta to B). This pair of equal, opposite, non-collinear forces constitutes a couple.

Torque of this couple = one Force x perpendicular distance between the forces

tau = F * (bsin(theta)) = (IaB) * (bsin(theta)) = I*(ab)Bsin(theta) = IABsin(theta), where A = a*b is the area of the loop.

For a coil of N turns, each carrying current I, the total torque is tau = NIABsin(theta).

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