# A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to the loop is parallel to a unit vector n^=−0.60i^−0.80j^. Calculate the magnitude of the torque on the loop due to the magnetic field.

The magnitude of the torque on the loop due to the magnetic field is .

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vector

We need to calculate the torque on the loop

Using formula of torque

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

Hence, The magnitude of the torque on the loop due to the magnetic field is .

The magnitude of the torque on the loop due to the magnetic field is 1.571 x 10⁻³ Nm.

Torque on the loop

The torque on the loop is calculated by using the following formulas;

τ = NIBAsinθ

where;

• N is the turn in the coil = -0.6i - 0.8j = √[(-0.6²) +(-0.8²)] = 1
• I is the current in the coil
• B is magnetic field
• A is area

A = πr²

A = πd²/4

A = π(0.1)²/4 = 7.855 x 10⁻³ m²

τ = 1 x 0.2 x (7.855 x 10⁻³) x sin90

τ = 1.571 x 10⁻³ Nm.

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