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

Answer 1

The Correct Option is A

To solve the problem of determining the torque experienced by a square loop in a magnetic field, we can use the formula for torque (\(\tau\)) on a current-carrying loop in a magnetic field: \(\tau = n \cdot I \cdot A \cdot B \cdot \sin\theta\). Here, \(n\) is the number of turns (1 for this loop), \(I\) is the current (10 A), \(A\) is the area of the loop, \(B\) is the magnetic field (0.2 T), and \(\theta\) is the angle between the magnetic field and the normal to the plane of the loop.

Since the magnetic field is parallel to the plane of the loop, the angle \(\theta\) between the magnetic field and the normal to the loop is 90 degrees. Thus, \(\sin\theta = \sin 90^\circ = 1\).

Now, calculate the area of the loop (\(A\)): For a square with side length 1 cm, convert this to meters to get 0.01 m. The area \(A = \text{side}^2 = (0.01)^2 = 0.0001 \, \text{m}^2\).

Substituting the values into the torque formula: \(\tau = 1 \cdot 10 \cdot 0.0001 \cdot 0.2 \cdot 1 = 2 \times 10^{-4} \, \text{Nm}\).

The torque experienced by the loop is therefore \(2 \times 10^{-4} \, \text{Nm}\).

Answer 2

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 1: 2 × 10^-4 Nm