Question

A solid cylinderical wire of radius 'R' carries a current T. The magnetic field is $5\mu T$ at a point, Whaich is '2R' distance away from the axis of wire. Magnetic field at a point which is R/3 distance inside from the surface of the wire is :-

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

Answer 1

The correct option is B $\frac{20}{3}\mu T$

$\text{Given:-}$ wire of radius=R

Curret=T

The magnetic field=5uT (At a distance 2R)

$\text{To find:-}$ Magnetic field at a point which is R/3 distance inside from the surface of the wire

$\text{Solution:-}$ In solid cylindrical conductor;

r>R then,

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

So B at 2R from axis

$S\mu T$= $\frac{{\mu }_{0}I}{2\pi \left(2R\right)}$

$\Rightarrow S\mu T=\frac{{\mu }_{0}I}{4\pi R}$ ............(1)

Now we have to take from the axis of cylinder. So,

r = R- $\frac{R}{3}=\frac{2R}{3}$

$B^{{}^{\prime }}=\frac{{\mu }_{0}I}{2}$

= $\frac{{\mu }_{0}Ir}{2\pi R^2}$

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

$B^{{}^{\prime }}=\frac{2{\mu }_{0}I}{6\pi R}$ .............(2)

Divide equation 1 by equation 2

$\frac{S\mu T}{B^{{}^{\prime }}}=\frac{\frac{{\mu }_{0}I}{4\pi R}}{\frac{2{\mu }_{0}I}{6\pi R}}$

$\Rightarrow \frac{S\mu T}{B^{{}^{\prime }}}=\frac{3}{4}$

$\therefore B^{{}^{\prime }}=\frac{20}{3}{\mu }_{0}T$

$\text{Hence the correct option is B}$