Question

What current must be maintained in a square loop (50 cm on a side) to create a torque of 1.0 N.m about an axis through its center and parallel to one of its sides when a magnetic field of magnitude 70 mT is directed at 40° to the plane of the loop!​

Answer 1

Answer:

70-40 =30. minius

50-30=20.

Answer 2

Concept:

• Magnetic fields produced by current carrying loops
• Torque produced by magnetic fields
• Torque = IAB sin θ

Given:

• Side of square loop = 50 cm = 0.5m
• Area of square loop A = 0.5^2 = 0.25 m^2
• Torque = 1.0 N m
• Magnetic field B = 70 mT = 70 *10^-3 = 7*10^-2 T
• Angle between loop and and magnetic field θ = 40°

Find:

• The value of the current carried in the loop

Solution:

We know that torque = IAB sin θ

1.0 N m = I * 0.25 m^2 * 7*10^-2 T sin 40°

I = 1.0/0.25*7*10^-2 * sin 40°

I = 89 A

The current to be maintained is  89 A.

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