Question

derive expression for torque acting on a plane coil in a magnetic field​

Answer 1

Explanation:

The angle between the direction of B and the vector perpendicular to the plane of the coil is 0. Let, a rectangular loop carrying a steady current I, placed in a uniform magnetic field B experiences a torque. Force on arm BC and DA are equal and opposite, and is acting along the axis of the coil. ...

Answer 2

Answer:

Let II = current flowing through the coil PQRSPQRS

a,ba,b = sides of the coil PQRSPQRS

A=abA=ab = area of the coil

θθ = angle between the direction of

B

and normal to the plane of the coil.

According to Fleming's left hand rule, the magnetic forces on sides PSPS and QRQR are equal, opposite and collinear (along the axis of the loop), so their resultant is zero.

The side PQPQ experiences a normal inward force equal to IbBIbB while the side RSRS experiences an equal normal outward force. These two forces form a couple which exerts a torque given by

τ=τ= Force ×× perpendicular distance

=IbB×asinθ=IbB×asin⁡θ

=IBAsinθ=IBAsin⁡θ

If the rectangular loop has NN turns, the torque increases NN times i.e.,

τ=τ= NIBAsinθNIBAsin⁡θ

But NIA=m,NIA=m, the magnetic moment of the loop, so

τ=mBsinθτ=mBsin⁡θ

In vector notation, the torque

τ

τ=

m

×

B

The direction of the torque τ is such that it rotates the loop clockwise about the axis of suspension.

Explanation:

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