Question

A cylindrical conductor AB of non-uniform area of cross-section carries a current of 5 A. The radius of the conductor at one end A is 0.5 cm. The current density at the other end of the conductor is half of the value at A The radius of the conductor at the end B is nearly​

Answer 1

Answer:

Magnetic field at a distance r inside a conductor of radius R is given by ampere’s circuital law

B×2πr=μ

0

Jπr

2

B=

2

μ

0

Jr

B=

2πR

2

μ

0

rI

Thus magnetic field inside a conductor is given by

B=

2πR

2

μ

0

rI

Magnetic field strength

H=

μ

0

B

H=

2πR

2

Ir

Explanation:

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