First slide

Magnetic field due to a straight current carrying wire

Question

A long straight wire of radius $a$ carries a steady current $I$ The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B^{'}$ , at radial distances $\frac{a}{2}$ and $2 a$ respectively, from the axis of the wire is

Moderate

A

1
B

4
C

$\frac{1}{4}$
D

$\frac{1}{2}$

Solution

Magnetic field at a point inside the wire at distance
$r (= \frac{a}{2}) from the axis of wire is$
$B = \frac{μ_{0} I}{2 π a^{2}} r = \frac{μ_{0} I}{2 π a^{2}} \times \frac{a}{2} = \frac{μ_{0} I}{4 π a}$
$M a g n e t i c f i e l d a t a p o i n t o u t s i d e t h e w i r e a t d i s \tan c e r (= 2 a) f r o m t h e a x i s o f w i r e i s$
$B^{'} = \frac{μ_{0} I}{2 π r} = \frac{μ_{0} I}{2 π} \times \frac{1}{2 a} = \frac{μ_{0} I}{4 π a} ∴ \frac{B}{B^{'}} = 1$

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