Question

A solid cylinder wire of radius $R$ carries a current $I$. The magnetic field is $5\mu \text{T}$ at a point, which is $2R$ distance away from the axis of wire. Magnetic field at a point which is $\frac{R}{3}$ distance inside from the surface of the wire is

• A.
  $\frac{10}{3}\mu \text{T}$
• B.
  $\frac{20}{3}\mu \text{T}$
• C.
  $\frac{5}{3}\mu \text{T}$
• D.
  $\frac{40}{3}\mu \text{T}$

Answer 1

The correct option is B

$\frac{20}{3}\mu \text{T}$
The magnetic field outside the solid wire at distance $r$ is,

$B_{outside}=\frac{{\mu }_{0}I}{2\pi r}=5\mu \text{T}$ (Given)

Here, $r=2R$

$B_{outside}=\frac{{\mu }_{0}I}{4\pi R}=5\mu \text{T}......\left(1\right)$

For magnetic field at distance $r$ inside the wire is,

$B_{inside}=\frac{{\mu }_{0}I}{2\pi R^2}r$

Here, $r=(R-\frac{R}{3})=\frac{2R}{3}$

$B_{inside}=\frac{{\mu }_{0}I}{2\pi R^2}\times \frac{2R}{3}=\frac{{\mu }_{0}I}{3\pi R}......\left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$,

$\frac{B_{inside}}{B_{outside}}=\frac{\frac{{\mu }_{0}I}{3\pi R}}{\frac{{\mu }_{0}I}{4\pi R}}=\frac{4}{3}$

$\Rightarrow B_{inside}=\frac{4}{3}\times B_{outside}=\frac{4}{3}\times 5\mu \text{T}=\frac{20}{3}\mu \text{T}$

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