Question

A long, straight wire with 2 A current flowing through it produces magnetic field strength 1 T at its surface. If the wire has a radius R, where within the wire is the field strength equal to 84 % of the field strength at the surface of the wire? Assume that the current density is uniform throughout the wire. (μ 0 = 4π × 10-7 T · m/A)

Answer 1

Answer:

$r = 3.36 \times 10^{-7} m$

Explanation:

As per Ampere's law of magnetic field we know that

line integral of magnetic field along closed ampere's loop is equal to the product of current enclosed and magnetic permeability of medium

So it is given as

$\int B. dl = \mu_0 i_{en}$

here we can say that enclosed current is given as

$i_{en} = \frac{i}{\pi R^2} (\pi r^2)$

now from ampere'e loop law for any point inside the wire we will have

$B.(2\pi r) = \mu_o (\frac{ir^2}{R^2}$

$B = \frac{\mu_0 i r}{2\pi R^2}$

now we know that magnetic field inside the wire is 84% of the field at its surface

so we will have

$0.84 \frac{\mu_o i}{2\pi R} = \frac{\mu_o i r}{2\pi R^2}$

so we have

$r = 0.84 R$

now we know

$\frac{\mu_o i}{2\pi R} = 1$

here i = 2 A

$R = 2\times 10^{-7} m$

so now we have

$r = 3.36 \times 10^{-7} m$