Question

A cylindrical conductor of radius R carries a current ′i′. The value of magnetic field at a point which is R/4 distance inside from the surface is 10 T. Find the value of magnetic field at point which is 4R distance outside from the surface.

• A. $\frac{80}{3}T$
• B. $\frac{4}{3}T$
• C. $\frac{40}{3}T$
• D. $\frac{8}{3}T$

Answer 1

The correct option is D $\frac{8}{3}T$

Magnetic field for an inside point of a solid cylindrical conductor, (r < R)

$B_{in}=\frac{{\mu }_{0}ir}{2\pi R^2}$

Where,
r = distance of point form axis,
R = radius of cylindrical conductor

Magnetic field for an outside point of solid cylindrical conductor , $(r^{\prime }>R)$

$B_{out}=\frac{{\mu }_{0}i}{2\pi r^{\prime }}$

Where,
r' = distance of point from axis

Dividing both equations, we get
$\frac{B_{in}}{B_{out}}=\frac{r{r}^{\prime }}{R^2}....\left(1\right)$

Inside point lies at R/4 distance from surface, so,
$r=R-\frac{R}{4}=\frac{3R}{4}$

Inside point magnetic field is given as,
$B_{in}=10T$

Outside point lies at distance 4R from surface , So,
$r^{{}^{\prime }}=R+4R=R$

Substituting the values in (1), we get

$\Rightarrow \frac{10}{B_{out}}=\frac{\left(\frac{3R}{4}\right)\left(5R\right)}{R^2}$

$\Rightarrow B_{out}=\frac{8}{3}T$

Hence, correct option is (B).