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Question: A cylindrical conductor of radius '\(R\)' carries a current '\(i\)'. The value of the magnetic field...

A cylindrical conductor of radius 'RR' carries a current 'ii'. The value of the magnetic field at a point which is R4\dfrac{R}{4} distance inside from the surface is 10T10\,T. Find the value of magnetic field at point which is 4R4R distance outside from the surface
A. 43 T\dfrac{4}{3}{\text{ T}}
B. 83 T\dfrac{8}{3}{\text{ T}}
C. 403 T\dfrac{{40}}{3}{\text{ T}}
D. 803 T\dfrac{{80}}{3}{\text{ T}}

Explanation

Solution

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<Rr < R
Binside=μ0ir2πR2{B_{inside}} = \dfrac{{{\mu _0}ir}}{{2\pi {R^2}}}
For r>Rr > R
Boutside=μ0i2πr{B_{outside}} = \dfrac{{{\mu _0}i}}{{2\pi r}}
This is ampere circuital law for inside and outside.Here BB represent magnetic field, dldl represents small length element, μ0{\mu _0} represent permeability of free space and ii represent current which produces magnetic field.

Complete step by step answer:
It is given that a cylindrical conductor of radius RR carries a current II. Value of magnetic field at a point R4\dfrac{R}{4} inside the cylindrical conductor of radius RR is equal to 10T10\,T. We have to find the value of magnetic field at a point 4R4R outside the cylindrical conductor of radius RR. So we can see that R4<R<4R\dfrac{R}{4} < R < 4R.
Applying the formula for R4<R\dfrac{R}{4} < R inside the conductor
Binside=μ0ir2πR2{B_{inside}} = \dfrac{{{\mu _0}ir}}{{2\pi {R^2}}}
Here r is equal to RR4R - \dfrac{R}{4}
Binside=μ0i(RR4)2πR2{B_{inside}} = \dfrac{{{\mu _0}i(R - \dfrac{R}{4})}}{{2\pi {R^2}}}

As given in the question Binside=10 T{B_{inside}} = 10{\text{ }}T
Therefore 10=μ0i(RR4)2πR210 = \dfrac{{{\mu _0}i(R - \dfrac{R}{4})}}{{2\pi {R^2}}}
Applying the formula for R<4RR < 4R inside the conductor
Boutside=μ0i2πr{B_{outside}} = \dfrac{{{\mu _0}i}}{{2\pi r}}
Here r is equal to R+4RR + 4R that is 5R
Boutside=μ0i2π(5R){B_{outside}} = \dfrac{{{\mu _0}i}}{{2\pi \left( {5R} \right)}}
In order to find out Boutside{B_{outside}} divide Binside{B_{inside}} by Boutside{B_{outside}}
BinsideBoutside=μ0i(RR4)2πR2μ0i2π(5R)\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 Boutside{B_{outside}}
10Boutside=(RR4)R21(5R)\dfrac{{10}}{{{B_{outside}}}} = \dfrac{{\dfrac{{(R - \dfrac{R}{4})}}{{{R^2}}}}}{{\dfrac{1}{{\left( {5R} \right)}}}}

As we can see that RR can be cut out from the equation which makes the equation simple
Here is an equation without RR.
10Boutside=(114)11(5)\dfrac{{10}}{{{B_{outside}}}} = \dfrac{{\dfrac{{(1 - \dfrac{1}{4})}}{1}}}{{\dfrac{1}{{\left( 5 \right)}}}}
Simplifying the equation
10Boutside=5×34\dfrac{{10}}{{{B_{outside}}}} = \dfrac{{5 \times 3}}{4}
Now taking the reciprocal to solve for Boutside{B_{outside}}
Boutside10=415\dfrac{{{B_{outside}}}}{{10}} = \dfrac{4}{{15}}
Taking ten on other side
Boutside=83\therefore {B_{outside}} = \dfrac{8}{3}
Hence the answer for Boutside{B_{outside}} is 83T\dfrac{8}{3}T.

Therefore the correct option is B.

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