Question

Magnetic field $40\text{cm}$ away from long straight wire carrying current $2\text{A}$ is $1.00\mu \text{T}$. At what distance, the magnetic field becomes $0.100\mu \text{T}$ ?

• A. $100\text{cm}$ away from wire
• B. $40\text{m}$ away from wire
• C. $4\text{cm}$ away from wire
• D. $4\text{m}$ away from wire

Answer 1

The correct option is D $4\text{m}$ away from wire

Magnetic field due to a long straight wire at a distance $d$ is given by,

$B=\frac{{\mu }_{0}I}{2\pi d}$

Given, $I=2\text{A},d_1=40\text{cm},B_1=1\mu \text{T}$

$1\times {10}^{-6}=\frac{{\mu }_0\left(2\right)}{2\pi (40\times {10}^{-2})}......\left(1\right)$

Now,

$B_2=0.1\mu \text{T}$, $d_2=?$, $I=2\text{A}$

$0.1\times {10}^{-6}=\frac{{\mu }_0}{2\pi }\frac{\left(2\right)}{d_2}.......\left(2\right)$

Dividing $\left(1\right)\text{by}\left(2\right)$, we get

$10=\frac{d_2}{40\times {10}^{-2}}$

$d_2=400\times {10}^{-2}\text{m}=4\text{m}$

$\therefore$ option $\left(d\right)$ is the correct answer.

Alternate solution:

$B=\frac{{\mu }_{0}I}{2\pi d}$

$\Rightarrow B\propto \frac{1}{d}$

$Bd=$ constant

$B_1{d}_1=B_2{d}_2$

Why this question ? Key point: The magnetic field due to long current carrying wire is given by, $B=\frac{{\mu }_{0}I}{2\pi d}$