Question

What is the magnetic field at the centre of a coil of
radius 11 cm and 100 turns
carrying a current of
2.1 A?
Ans= 1.2 × 10^-3​

Answer 1

Answer:

Here's your explanation....

Explanation:

Number of turns on the circular coil, n = 100  

Radius of each turn, r = 11.0 cm = 0.11 m

Current flowing in the coil, I = 2.1 A

Magnitude of the magnetic field at the center of the coil is given by the relation,

| B | = μ0 2πnI / 4π r

Where,

μ0 = Permeability of free space = 4π × 10⁻⁷ T m A⁻¹

| B | = 4π x 10⁻⁷ x 2π x 100 x 2.1 / 4π x 0.11

      = 1.2 x 10⁻³ T

Hence, the magnitude of the magnetic field is 1.2 x 10⁻³T.

Answer 2

Explanation:

thanks for free points .............................