Question
Question: A cylindrical conductor of radius '\(R\)' carries a current '\(i\)'. The value of the magnetic field...
A cylindrical conductor of radius 'R' carries a current 'i'. The value of the magnetic field at a point which is 4R distance inside from the surface is 10T. Find the value of magnetic field at point which is 4R distance outside from the surface
A. 34 T
B. 38 T
C. 340 T
D. 380 T
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<R
Binside=2πR2μ0ir
For r>R
Boutside=2πrμ0i
This is ampere circuital law for inside and outside.Here B represent magnetic field, dl represents small length element, μ0 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 4R inside the cylindrical conductor of radius R is equal to 10T. 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 4R<R<4R.
Applying the formula for 4R<R inside the conductor
Binside=2πR2μ0ir
Here r is equal to R−4R
Binside=2πR2μ0i(R−4R)
As given in the question Binside=10 T
Therefore 10=2πR2μ0i(R−4R)
Applying the formula for R<4R inside the conductor
Boutside=2πrμ0i
Here r is equal to R+4R that is 5R
Boutside=2π(5R)μ0i
In order to find out Boutside divide Binside by Boutside
BoutsideBinside=2π(5R)μ0i2πR2μ0i(R−4R)
After cutting down the same variables and inserting the value of Boutside
Boutside10=(5R)1R2(R−4R)
As we can see that R can be cut out from the equation which makes the equation simple
Here is an equation without R.
Boutside10=(5)11(1−41)
Simplifying the equation
Boutside10=45×3
Now taking the reciprocal to solve for Boutside
10Boutside=154
Taking ten on other side
∴Boutside=38
Hence the answer for Boutside is 38T.
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 Binside and they also not add the value of r in R when we find Boutside. Here we are referring to R as the radius of a given metal cylinder while r as the point where magnetic field is asked.