Question

A long wire is known to have a radius greater than 5.6 mm and to carry a current that is uniformly distributed over its cross section. The magnitude of the magnetic field due to that current is 0.44 mT at a point 4.0 mm from the axis of the wire, and 0.27 mT at a point 32 mm from the axis of the wire. What is the radius of the wire?

Answer 1

The radius is
`R=8.86mm`

Step-by-step explanation:

Generally Magnetic field due to a current carrying wire is is mathematically represented as

`B = (\mu_0 I )/(2 \pi r_1)`

Where
`r_1` is the distance from the axis of the wire to the first point we are considering

I is the current

`\mu_0` is the permeability of free space

Making I the subject

`I = (2 \pi r_1 B)/(\mu_0)`

Generally Magnetic field inside a current carrying wire is is mathematically represented as

`B_0=(\mu_0 I r)/(2 \pi R^2)`

Now the r here is the distance from the axis of the wire to the second point we are considering

And R i the radius of the wire

Making R the subject of the formula

`R = \sqrt{(\mu_0 I r)/(2 \pi B_0) }`

Substituting for I we have

`R = \sqrt{(\mu_0 ((2 \pi r_1 B)/(\mu_0) ) r)/(2 \pi B_0) }`

`R = \sqrt{(B)/(B_0) r * r_1}`

Where B = 0.27mT

And
`B_0` = 0.44mT

r = 32mm

`r_1` = 4.0mm

Substituting values

`R = \sqrt{(0.27)/(0.44) *4 *32 }`

`R=8.86mm`