A rectangular coil comprising 100 turns has a length of 5 cm and a width of 4 cm . It is positioned such that its plane is parallel to a uniform magnetic field, and a current of 2 A is passed through the coil. Determine

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Correct answer: D. 0.5 T

The torque acting on a current-carrying coil in a uniform magnetic field is given by:

\tau = N I A B \sin\theta

Given that the plane of the coil is parallel to the magnetic field, the angle between the area vector and the magnetic field is

\theta = 90^{}^{{}^{\circ}}

.

Substituting the given values:

N = 100

I = 2 \text{ A}

A = 5 \text{ cm} \times 4 \text{ cm} = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2

\tau = 0.2 \text{ N m}

We get:

0.2 = 100 \times 2 \times 20 \times 10^{-4} \times B \times \sin(90^{}^{{}^{\circ}})

0.2 = 4000 \times 10^{-4} \times B

0.2 = 0.4 \times B

B = \dfrac{0.2}{0.4} = 0.5 \text{ T}

A rectangular coil comprising 100 turns has a length of 5 cm and a...

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