Question

A long straight wire carries a conventional current of 0.7 A. What is the approximate magnitude of the magnetic field at a location a perpendicular distance of 0.053 m from the wire due to the current in the wire

Answer 1

Answer:

2.64 x 10⁻⁶T

Explanation:

The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;

B = (μ₀ I) / (2π r)                ----------------(i)

B is magnetic field

I is current through the wire

r is the distance from the wire

μ₀  is the magnetic constant = 4π x 10⁻⁷Hm⁻¹

From the question;

I = 0.7A

r = 0.053m

Substitute these values into equation (i) as follows;

B =  (4π x 10⁻⁷ x 0.7) / (2π x 0.053)

B = 2.64 x 10⁻⁶T

Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T