First slide

Ampere's circuital law

Question

Magnetic induction on the surface of a long current carrying wire of radius R is 8 µT. Then magnetic induction at a distance R/4 from the axis of the conductor is

Moderate

A

4 µT
B

3 µT
C

2 µT
D

1 µT

Solution

On the surface, $B_{1} = \frac{μ_{o} i}{2 πR}$ .
At a distance r (r ≤ R) from the axis of the conduction;
$B_{2} = \frac{1}{2 π} \frac{μ_{o} jr}{R^{2}} .$
At r = R/4, $B_{2} = \frac{μ_{o} i}{2 {πR}^{2}} . \frac{R}{4} = \frac{1}{4} (\frac{μ_{o} i}{2 πR}) = \frac{B_{1}}{4} = \frac{8}{4} = 2 μT$ .

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