Question

tic field 2 cm from the center of the wire? A 40 cm wire with a radius of 3 cm is oriented along the y axis and carries a current of 2 A. What is the magnitude of the magnetic field 4 cm from the center of the wire

Answer 1

Answer:

a) B = 1.99 x 10⁻⁴ Tesla

b) B = 0.88 x 10⁻⁴ Tesla

Explanation:

According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:

B = μ₀ I L/4πr²

where,

μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹

I = current = 2 A

L = Length of wire = 40 cm = 0.4 m

a)

r = radius of magnetic field = 2 cm = 0.02 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²

B = 1.99 x 10⁻⁴ Tesla

b)

r = radius of magnetic field = 3 cm = 0.03 m

Therefore,

B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²

B = 0.88 x 10⁻⁴ Tesla