First slide

Magnetic field due to current element

Question

A long cylindrical wire of radius ‘a’ carries a current i distributed uniformly over its cross section. If the magnetic fields at distances r<a and R>a from the axis have equal magnitude, then

Moderate

A

$a = \frac{R + r}{2}$
B

$a = \sqrt{R r}$
C

$a = R r / R + r$
D

$a = R^{2} / r$

Solution

Applying Ampere’s circular law.
$I n s i d e p o i n t, B 2 π r = \frac{μ_{0} I π r^{2}}{π a^{2}}$ for $r < a$
And $O u t s i d e p o i n t, B 2 π R = μ_{0} I$ for $R > a$
Taking the ratio $\frac{r}{R} = \frac{r^{2}}{a^{2}} \Rightarrow a = \sqrt{R r}$ .

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