Question

A long straight solid conductor of radius $8cm$ carries a current of $2A$, which is uniformly distribution over its circular cross-section. Find the magnetic field induction at a distance of $3cm$ from the axis of the conductor.

Answer 1

Given that,

Radius $r=8cm$

Current $I=2A$

Distance $d=3cm$

Now, the magnetic field at a point inside the conductor

$B=\frac{{\mu }_{0}Id}{2\pi r^2}$

$B=\frac{4\pi \times {10}^{-7}\times 2\times 3\times {10}^{-2}}{2\pi \times 64\times {10}^{-4}}$

$B=0.1875\times {10}^{-5}$

$B=1.88\times {10}^{-6}T$

Hence, the magnetic field is $1.88\times {10}^{-6}T$