Answer:
The current the wire must carry is 1250 A
Explanation:
Given;
strength of magnetic field, B = 5.0 × 10⁻⁴ T
distance from the wire, r = 0.50 m
The strength of magnetic field on the current carrying wire is given as;
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
where;
B is the magnetic field strength
μ₀ is permeability of free space, = 4π x 10⁻⁷ T.m/A
I is the current on the wire
r is the distance from the wire
Make current "I" the subject of the formula;
[tex]B = \frac{\mu_o I}{2\pi r} \\\\I = \frac{2\pi rB}{\mu_o} \\\\I = \frac{2\pi *0.5*5*10^{-4}}{4\pi *10^{-7} }\\\\I = 1.25 *10^3 \ A\\\\I = 1250 \ A[/tex]
Therefore, the current the wire must carry is 1250 A
