Question

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.50 x 10-4 T) at a distance of 15 cm from the wire, what is the maximum current the wire can carry? Express your answer using two significant figures.

Answer 1

Answer:

$I = 37.5\ A$

Explanation:

Given,

Magnetic field,B = 0.5 x 10⁻⁴ T

distance,r= 15 cm = 0.15 m

Current = ?

Using Ampere's law of magnetic field

$B = \dfrac{\mu_0I}{2\pi r}$

$I= \dfrac{B (2\pi r)}{\mu_0}$

$I= \dfrac{0.5\times 10^{-4}\times (2\pi \times 0.15)}{4\pi \times 10^{-7}}$

$I = 37.5\ A$

Current in the wire is equal to $I = 37.5\ A$

Answer 2

The maximum current this wire can carry is equal to 37.5 Amperes.

Given the following data:

• Magnetic field = $0.5 \times 10^{-4}\;T$.

• Distance = 15 cm to m = 0.15 meter.

Scientific data:

• Permeability of free space = $4\pi \times 10^{-7}$

How to calculate the maximum current.

In order to determine the maximum current, we would apply Ampere's law of magnetic field.

Mathematically, Ampere's law of magnetic field is given by this formula:

$I=\frac{2B\pi r}{\mu_o }$

Where:

• B is the magnetic field.

• I is the current.

• r is the distance.

• $\mu_o$ is the permeability of free space.

Substituting the given parameters into the formula, we have;

$I=\frac{2 \pi \times 0.5 \times 10^{-4}\times 0.15}{4\pi \times 10^{-7} }\\\\I=\frac{0.5 \times 10^{-4}\times 0.15}{2\pi \times 10^{-7} }$

I = 37.5 Amperes.

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