A single-turn circular loop carries a current of 5.00 A . Determine the radius of the loop if the magnetic field B at its centre is 0.200 mT .

Correct answer: B. 1.57 cm The magnetic field at the centre of a circular loop is given by B = (μ_0 I)/(2R) Rearranging for the radius R , we get R = (μ_0 I)/(2B) Substituting the given values, I = 5.00 A and B = 0.200 mT = 2.00 × 10^ -4 T : R = 4π × 10^ -7 × 5.00 2 × 2.00 × 10^ -4 R = 20π × 10^ -7 4.00 × 10^ -4 R = 5π × 10^ -3 m R = 15.7 × 10^ -3 m = 1.57 cm

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Magnetic Field due to a Current

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Magnetic Field due to a Current

A single-turn circular loop carries a current of 5.00 A . Determine...

A single-turn circular loop carries a current of

5.00 \text{ A}

. Determine the radius of the loop if the magnetic field

B

at its centre is

0.200 \text{ mT}

Options

A $0.785 \text{ cm}$
B $1.57 \text{ cm}$
C $3.14 \text{ cm}$
D $15.7 \text{ cm}$

Correct answer

B. $1.57 \text{ cm}$

Step-by-step solution

The magnetic field at the centre of a circular loop is given by

B = \dfrac{\mu_0 I}{2R}

Rearranging for the radius

R

, we get

R = \dfrac{\mu_0 I}{2B}

Substituting the given values,

I = 5.00 \text{ A}

and

B = 0.200 \text{ mT} = 2.00 \times 10^{-4} \text{ T}

R = \dfrac{4\pi \times 10^{-7} \times 5.00}{2 \times 2.00 \times 10^{-4}}

R = \dfrac{20\pi \times 10^{-7}}{4.00 \times 10^{-4}}

R = 5\pi \times 10^{-3} \text{ m}

R = 15.7 \times 10^{-3} \text{ m} = 1.57 \text{ cm}

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