Question

A hollow cylindrical conductor of inner radius 0.0115 m and outer radius 0.0299 m carries a uniform current of 2.31 A. What is the magnitude of the magnetic field at a radius of 0.0169 m?

a) 1.42 × 10^(-4) T
b) 2.05 × 10^(-4) T
c) 2.92 × 10^(-4) T
d) 3.71 × 10^(-4) T

Answer 1

To find the magnetic field at a radius of 0.0169 m, we can use Ampere's Law. The magnitude of the magnetic field is approximately 2.05 × 10^(-4) T.

Step-by-step explanation:

To find the magnetic field at a radius of 0.0169 m, we can use Ampere's Law. Ampere's Law states that the magnetic field around a closed loop is proportional to the current passing through the loop.

In our case, the closed loop is a circular path with a radius of 0.0169 m. The current passing through this loop is the total current passing through the conductor, which is given as 2.31 A. Therefore, we can calculate the magnetic field using the formula: B = (μ₀ * I) / (2 * π * r), where μ₀ is the permeability of free space, I is the current, and r is the radius of the loop.

Plugging in the values, we get: B = (4π * 10^(-7) T m/A * 2.31 A) / (2 * π * 0.0169 m) = 2.0586 * 10^(-4) T.

Therefore, the magnitude of the magnetic field at a radius of 0.0169 m is approximately 2.05 × 10^(-4) T.