Question

A copper wire of diameter $2.4\text{mm}$ carries a current $i$. The maximum magnetic field due to this wire $5\times {10}^{-4}\text{T}$, then the value of $i$ is

• A. $12\text{A}$
• B. $4\text{A}$
• C. $3\text{A}$
• D. $5\text{A}$

Answer 1

The correct option is C $3\text{A}$

We know for a solid current carrying cylindrical wire, the maximum value of magnaetic field is obtained at a distance $r=R$

$B_{max}=\frac{{\mu }_{0}i}{2\pi R}$


$i=\frac{B_{max}\left(2\pi R\right)}{{\mu }_0}$

$i=\frac{5\times {10}^{-4}\times 2\pi \times 1.2\times {10}^{-3}}{4\pi \times {10}^{-7}}$

$i=3\text{A}$

Why this Question ? Key point: For, $d=R$, $B_{max}=\frac{{\mu }_{0}I}{2\pi R}$