Question

A circular loop carrying a current of 2.0 A is in a magnetic field of 3.5 T. The loop has an area of 0.12 m2 and its plane is oriented at a 37° angle to the field. What is the magnitude of the magnetic torque on the loop?

Answer 1

Answer:

Torque,  

Explanation:

It is given that,

Magnetic field, B = 3.5 T

Current, I = 2 A

Area of cross section, A = 0.12 m²

Angle between the plane and the magnetic field, θ = 37°

We need to find the magnitude of torque on the loop. It is given by :

$\tau=IAB\ sin\theta$

Here, $\theta$ is the angle between the magnetic field and the normal to the coil.

$\tau=2\ A\times 0.12\ m^2\times 3.5\ T\ sin(90-37)$

$\tau=0.67\ N-m$

So, the magnitude of the magnetic torque on the loop is 0.67 N-m. Hence, this is the required solution.

Answer 2

The magnitude of the magnetic torque on the loop is 5.59 N.

To calculate the magnitude of the magnetic torque in the loop, we use the formula below.

Formula:

• τ = BIAsinθ............. Equation 1

Where:

• τ = Magnetic torque
• B = Magnetic field
• I = Current
• A = Area of the loop
• θ = Angle.

From the question,

Given:

• B = 3.5 T
• A = 0.12 m²
• I = 2 A
• θ = (90-37 ) = 53°

Substitute these values into equation 1

• τ = 3.5×2×0.12×sin53°
• τ = 5.59 N

Hence, the magnitude of the magnetic torque on the loop is 5.59 N.

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