A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is ________ A -m². [Given π = 3.14 ] Numerical answer type. Work it out, then open the answer.

Answer: 11. 2π r = 3.14 m ⇒ r = 0.5 m. A = π r² = 3.14×0.25 = 0.785 m ² M = IA = 14×0.785 = 10.99 So M ≈ 11 A m ².

A wire of length 314 cm carrying current of 14 A is bent to form a… | JEE Main 2022 (29 Jul, Shift 1)

Q48JEE Main 2022 · 29 JUL · SHIFT 1Medium

Current Loop as a Magnetic Dipole

A wire of length $314\text{ cm}$ carrying current of $14\text{ A}$ is bent to form a circle. The magnetic moment of the coil is ________ A $-\text{m}^{2}$ . [Given $\pi = 3.14$ ]

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 11

$2\pi r = 3.14$ m $\Rightarrow r = 0.5$ m.
$A = \pi r^2 = 3.14\times0.25 = 0.785$ m $^2$
$M = IA = 14\times0.785 = 10.99$
So $M \approx 11$ A m $^2$ .

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Current Loop as a Magnetic Dipole question from JEE Main 2022 (29 Jul, Shift 1)

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