Question

A long solid cylindrical conductor of radius $10\text{cm}$ carries a current of $4\text{A}$ which is uniformly distributed over its circular cross-section. The magnetic field at a distance of $5\text{cm}$ from the axis of the conductor is

• A. $4\times {10}^{-6}\text{T}$
• B. $2\times {10}^{-8}\text{T}$
• C. $0.8\times {10}^{-7}\text{T}$
• D. $0.4\times {10}^{-8}\text{T}$

Answer 1

The correct option is A $4\times {10}^{-6}\text{T}$

Using Amperes law, we know that for any inside point magnaetic field is given by,

$B=\frac{{\mu }_{0}Ir}{2\pi R^2}$

$B=\frac{4\pi \times {10}^{-7}\times 4\times 5\times {10}^{-2}}{2\pi \times (10\times {10}^{-2}{)}^2}$

$B=4\times {10}^{-6}\text{T}$

Hence, option $\left(a\right)$ is the correct answer.