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Question
Derive the expression for the torque on a current-carrying coil in a magnetic field.
Solution
Consider a rectangular loop PQRS carrying current I is placed in a uniform magnetic field B. Let a and b be the length and breadth rectangular loop respectively. The unit vector n̂ normal to the plane of the loop makes an angle θ with the magnetic field.
Rectangular coil placed in a magnetic field
- The magnitude of the magnetic force acting on the current-carrying arm PQ is FPQ = IaBsin (π/2) = IaB.
- The magnitude of the force on the arm QR is FQR =IbBsin `(pi/2 - theta)` = lbB cos θ and its direction.
- The magnitude of the force on the arm RS is FRS =IaBsin (π/2) = IaB and its direction is downwards.
- The magnitude of the force acting on the arm SP is FSP = lbB`(pi/2 - theta)` = lbB cos θ and its direction.
- Since the forces FQR and FSP are equal, opposite, and collinear, they cancel each other. But the forces FPQ and FRS.
Side view of current loop - The magnitude of torque acting on the arm PQ about AB is τPQ = (1/2 sinθ) and points in the direction of AB. The magnitude of the torque acting on the arm. RS about AB is τRS = (b/2 sinθ) IaB and points also in the same direction AB.
- The total torque acting on the entire loop about an axis AB is given by
`tau = ("b"/2 sin theta) "F"_"PQ" + ("b"/2 sin theta) "F"_"RS"`
= Ia(bsinθ)B
τ = IABsinθ along the direction B.
In vector form,
`vec tau= (vec"IA") xx vec"B"`. - The above equation can also be written in terms of magnetic dipole moment.
`vectau` = \[\overrightarrow{\mathrm{pm}}\] × \[\overrightarrow{\mathrm{B}}\] where \[\overrightarrow{\mathrm{pm}}\] = \[\overrightarrow{\mathrm{IA}}\] - If there are N turns in the rectangular loop, the torque is given by
τ = NIAB sinθ
Special cases:
- When θ =90° or the plane of the loop is parallel to the magnetic field, the torque on the current loop is maximum.
τmax = IAB - When θ = 0°/180° or the plane of the loop is perpendicular to the magnetic field, the torque on the current loop is zero.
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