Question

A cylindrical conductor of radius '$R$' carries a current '$i$'. The value of the magnetic field at a point which is $\dfrac{R}{4}$ 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. $\dfrac{4}{3}{\text{ T}}$
B. $\dfrac{8}{3}{\text{ T}}$
C. $\dfrac{{40}}{3}{\text{ T}}$
D. $\dfrac{{80}}{3}{\text{ T}}$

Answer 1

Hint: In order to solve this question, we have to use ampere circuital law to find the magnetic field use magnetic field of the region inside cylindrical conductor and use ampere circuital law for outside cylindrical conductor after then we have to divide inside and outside magnetic field to find the answer magnetic field outside cylinder.

Formula used:
For $r < R$
${B_{inside}} = \dfrac{{{\mu _0}ir}}{{2\pi {R^2}}}$
For $r > R$
${B_{outside}} = \dfrac{{{\mu _0}i}}{{2\pi r}}$
This is ampere circuital law for inside and outside.Here $B$ represent magnetic field, $dl$ represents small length element, ${\mu _0}$ represent permeability of free space and $i$ represent current which produces magnetic field.

Complete step by step answer:
It is given that a cylindrical conductor of radius $R$ carries a current $I$. Value of magnetic field at a point $\dfrac{R}{4}$ inside the cylindrical conductor of radius $R$ is equal to $10\,T$. We have to find the value of magnetic field at a point $4R$ outside the cylindrical conductor of radius $R$. So we can see that $\dfrac{R}{4} < R < 4R$.
Applying the formula for $\dfrac{R}{4} < R$ inside the conductor
${B_{inside}} = \dfrac{{{\mu _0}ir}}{{2\pi {R^2}}}$
Here r is equal to $R - \dfrac{R}{4}$
${B_{inside}} = \dfrac{{{\mu _0}i(R - \dfrac{R}{4})}}{{2\pi {R^2}}}$

As given in the question ${B_{inside}} = 10{\text{ }}T$
Therefore $10 = \dfrac{{{\mu _0}i(R - \dfrac{R}{4})}}{{2\pi {R^2}}}$
Applying the formula for $R < 4R$ inside the conductor
${B_{outside}} = \dfrac{{{\mu _0}i}}{{2\pi r}}$
Here r is equal to $R + 4R$ that is 5R
${B_{outside}} = \dfrac{{{\mu _0}i}}{{2\pi \left( {5R} \right)}}$
In order to find out ${B_{outside}}$ divide ${B_{inside}}$ by ${B_{outside}}$
$\dfrac{{{B_{inside}}}}{{{B_{outside}}}} = \dfrac{{\dfrac{{{\mu _0}i(R - \dfrac{R}{4})}}{{2\pi {R^2}}}}}{{\dfrac{{{\mu _0}i}}{{2\pi \left( {5R} \right)}}}}$
After cutting down the same variables and inserting the value of ${B_{outside}}$
$\dfrac{{10}}{{{B_{outside}}}} = \dfrac{{\dfrac{{(R - \dfrac{R}{4})}}{{{R^2}}}}}{{\dfrac{1}{{\left( {5R} \right)}}}}$

As we can see that $R$ can be cut out from the equation which makes the equation simple
Here is an equation without $R$.
$\dfrac{{10}}{{{B_{outside}}}} = \dfrac{{\dfrac{{(1 - \dfrac{1}{4})}}{1}}}{{\dfrac{1}{{\left( 5 \right)}}}}$
Simplifying the equation
$\dfrac{{10}}{{{B_{outside}}}} = \dfrac{{5 \times 3}}{4}$
Now taking the reciprocal to solve for ${B_{outside}}$
$\dfrac{{{B_{outside}}}}{{10}} = \dfrac{4}{{15}}$
Taking ten on other side
$\therefore {B_{outside}} = \dfrac{8}{3}$
Hence the answer for ${B_{outside}}$ is $\dfrac{8}{3}T$.

Therefore the correct option is B.

Note: Many of the students mistake in inserting the value of $r$ in ampere circuit law equation they usually do not subtracting r from R when we find ${B_{inside}}$ and they also not add the value of $r$ in $R$ when we find ${B_{outside}}$. Here we are referring to $R$ as the radius of a given metal cylinder while $r$ as the point where magnetic field is asked.