Question

A square loop with each side 1 cm, carrying a current of 10 A, is placed in a magnetic field of 0.2 T. The direction of the magnetic field is parallel to the plane of the loop. The torque experienced by the loop is .
Fill in the blank with the correct answer from the options given below

• A. zero
• B. \(2 × 10^{-4} Nm\)
• C. \(2 × 10^{-2} Nm\)
• D. 2Nm

Answer 1

The Correct Option is A

To find the torque experienced by the loop, we'll use the formula:

τ = NIABsinθ

Where:

• τ is the torque
• N is the number of turns in the loop (N = 1 in this case)
• I is the current (I = 10 A)
• A is the area of the loop
• B is the magnetic field (B = 0.2 T)
• θ is the angle between the normal to the loop's plane and the magnetic field.

1. Area of the Loop:

The loop is a square with each side 1 cm (0.01 m).

A = (0.01 m)² = 10^-4 m²

2. Angle θ:

The magnetic field is parallel to the plane of the loop. This means the normal to the loop's plane is perpendicular to the magnetic field. Therefore, θ = 90°.

sin(90°) = 1

3. Calculate the Torque:

τ = NIABsinθ

τ = (1) * (10 A) * (10^-4 m²) * (0.2 T) * (1)

τ = 2 * 10^-4 Nm

Therefore, the torque experienced by the loop is 2 × 10^-4 Nm.

The correct answer is:

Option 2: 2 × 10^-4 Nm