Question

A circular coil of radius 1 m carries a current 2.5 A.
If it is placed in a magnetic field of 10 Wb/m2
The work done to rotate it from position to stable
equilibrium to unstable equilibrium is
0507 J
(2) 2.5 J
(3) 25 J
(4) 2.5 J
​

Answer 1

Change in P.E= WORK DONE=MB-(-MB)=2MB=2IAB

Put all the values you will get the answer.

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Answer 2

Answer:

A circular coil of radius 1 m carrying current 2.5A, placed in magnetic field 10Wb/m². Work done in rotating coil from stable to unstable equilibrium is 157.07 J.

Explanation:

• A circular current carrying loop behaves as a magnetic dipole.
• The magnetic dipole moment, m of any current carrying loop is equal to the product of current and its loop area.

$magnetic \: dipole \: moment, m = IA$

• Its direction is defined to be normal to the plane of the loop in the sense given by right hand thumb rule.
• When a magnetic dipole is placed in a magnetic field, B it experience a torque.
• If the dipole is rotated against the action of this torque, work has to be done. This work is stored as potential energy of the dipole.

$potential \: energy \:, U = - mB cosθ$

• If θ = 0⁰, the magnetic dipole is in stable equilibrium.
• If θ = 180⁰, the magnetic dipole moment is in unstable equilibrium.

Given that:

• Radius of coil = 1 meter
• Current in coil = 2.5 A
• Magnetic field strength = 10 Wb/m²

Solution:

• Potential energy in stable equilibrium is -mB.
• Potential energy in unstable equilibrium is mB.
• Total change in potential energy = mB - (-mB) = 2mB.
• This potential energy is equal to work done.

$Work \: done, W = 2mB \\ = 2IAB \\ = 2 \times 2.5 \times π \times (1)² \times 10 \\ = 50π \\ = 157.07J$

• Hence, the total work done is 157.07 J.