Question

a circular current carrying loop of radius 25 cm having 12 turns placed in uniform magnetic field of 0.3T .The current in the loop is 5A . If its normal to the plane makes an angle of 60 with magmetic field fInd tOrque

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Answer 1

3.06 Nm is the required answer.

GIVEN:

• Radius of current carrying loop, r = 25 cm = 25 × 10^-2 m

• Number of turns, n = 12

• Magnetic field , B = 0.3 T

• Current in the loop, I = 5 A

• Angle , θ = 60°

TO FIND:

Torque

FORMULAE:

• Area of circular loop , A = π r²

• Magnetic moment, M = nIA

• Torque = MB sin θ

• sin 60° = √3/2

SOLUTION:

STEP 1: TO FIND MAGNETIC MOMENT:

M = nIA

M = nI π r²

$= 12 \times 5 \times 3.14 \times {(25 \times {10}^{ - 2} )}^{2} \\ \\ = 60 \times 3.14 \times 25 \times 25 \times {10}^{ - 4} \\ \\ = 117750 \times {10}^{ - 4} \\ \\ = \frac{117750}{10000} \\ \\ M = 11.775 \: A {m}^{2}$

Magnetic moment , M is 11.775 Am²

STEP 2: TO FIND TORQUE:

Torque = MB sin θ

$torque = 11.775 \times 0.3 \times \sin(60) \\ \\ = 11.775 \times 0.3 \times \frac{ \sqrt{3} }{2} \\ \\ = 11.775 \times 0.3 \times \frac{1.732}{2} \\ \\ = 11.775 \times 0.3 \times 0.866 \\ \\ torque = 3.059145 \: Nm \\ \\ torque = 3.06 \: Nm$

ANSWER:

Torque produced is 3.06 Nm

TORQUE:

• Torque is defined as the moment of force.

• Torque is the force which is responsible for the rotational movement.

• When a current carrying conductor is placed in uniform magnetic field, it experiences no force. But it experiences torque .

• When the current carrying conductor is placed in non uniform magnetic field, it experiences both force and torque.