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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 solid wire with a radius of 10 cm carries a 5.0 A current that is...

A solid wire with a radius of

10 \text{ cm}

carries a

5.0 \text{ A}

current that is uniformly distributed across its cross section. Determine the magnetic field

B

at a point located at a distance of

2 \text{ cm}

from the axis.

Options

A $4.0 \mu\text{T}$
B $0.2 \mu\text{T}$
C $2.0 \mu\text{T}$
D $50 \mu\text{T}$

Correct answer

C. $2.0 \mu\text{T}$

Step-by-step solution

Given:
Radius of the wire,

R = 10 \text{ cm} = 0.1 \text{ m}

Current,

I = 5.0 \text{ A}

Distance from the axis,

r = 2 \text{ cm} = 0.02 \text{ m}

The magnetic field inside a solid cylindrical wire carrying a uniform current at a distance

r < R

is given by Ampere's Law:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}

B(2\pi r) = \mu_0 \left( \dfrac{I}{\pi R^2} \right) \pi r^2

B = \dfrac{\mu_0 I r}{2\pi R^2}

Substituting the given values:

B = \dfrac{4\pi \times 10^{-7} \times 5.0 \times 0.02}{2\pi \times (0.1)^2}

B = \dfrac{2 \times 10^{-7} \times 0.1}{0.01}

B = 2 \times 10^{-6} \text{ T} = 2.0 \mu\text{T}

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