First slide

Ampere's circuital law

Question

A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a /2and 2a is

Moderate

A

1
B

$\frac{1}{2}$
C

$\frac{1}{3}$
D

4

Solution

The situation is shown in fig.
Using Ampere's circuital law
$(2 π \cdot \frac{a}{2}) (B)_{a / 2} = μ_{0} \times J \times π {(\frac{a}{2})}^{2}$ ………..(1)
and $(2 π . 2 a) (B)_{2 a} = μ_{0} \times J \times π (2 a)^{2}$ ………(2)
Dividing eq. (1) by eq. [2), we get
$\begin{array}{r} \frac{1}{4} \frac{(B)_{a / 2}}{(B)_{2 a}} = \frac{1}{16} \\ \frac{(B)_{a / 2}}{(B)_{2 a}} = \frac{1}{4} \end{array}$

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