Two long straight wires are parallel and 9.5 cm apart. They are to carry equal currents such that the magnetic field at a point halfway between them has magnitude 280 T.
(a) Should the currents be in the same or opposite directions?
(b) How much current is needed?

Respuesta :

Answer:

(a) the current will flow in opposite direction

(b) the current needed is 33.25 A

Explanation:

(a) At the center of the two parallel wires, the two wires will have the same magnitude of magnetic field. In order to have a non a zero value of magnetic field at the center, the field must be in the same direction and the current will flow in opposite direction according to right hand rule.

(b) How much current is needed

Given;

distance between the two parallel wires, d = 9.5 cm = 0.095 m

magnitude of magnetic field at a point halfway between the wires, [tex]B_c[/tex] = 280 μT (This unit was corrected to obtain feasible current)

The magnetic field at distance R due to an infinite wire is given by;

[tex]B = \frac{\mu_o I}{2\pi R}[/tex]

At the center of the wire, [tex]B_c = 2B[/tex]

[tex]B_c = 2(\frac{\mu_o I}{2\pi R} )\\\\B_c = \frac{\mu_o I}{\pi R}[/tex]

where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

R is the center point between the wires, R = d/2 = 0.095m / 2 = 0.0475 m

I is the current needed

[tex]B_c = \frac{\mu_o I}{\pi R} \\\\I = \frac{B_c \pi R}{\mu_o} \\\\I = \frac{280* 10^{-6}*\pi *0.0475}{4\pi *10^{-7}} \\\\I = 33.25 \ A[/tex]