Question

A current carrying coil of radius R is placed in x-y plane. If a uniform magnetic field is along +x direction in the same plane than find the torque on c o i l

Answer 1

A current carrying coil of radius R is placed in x-y plane, if a uniform magnetic field is along +x direction in the same plane then the Torque on coil in z direction.

Explanation:

let I = current flowing through the coil

    a,b = sides of the coil

   A = ab = Area of the coil

  Q = 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 parallel sides of coil are equal , opposite and collinear (along the axis of loop), so their resultant is zero.

the one side experience a normal inward force equal to IbB while the other parallel side experience an equal normal outward force . these two forces form a couple which exert a Torque

torque = Force x perpendicular distance

n           = IbB x asin Q = IBA sin Q

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

so     Torque = mB sin Q.

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

Answer 2

Answer:

πR2IBo (j)

Explanation:

hope this helps you