Question 1

# A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer

Number of turns on the circular coil, *n* = 100

Radius of each turn,* **r* = 8.0 cm = 0.08 m

Current flowing in the coil, *I* = 0.4 A

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

| B | = μ_{0} 2πnI / 4π r

Where,

μ_{0 }= Permeability of free space

= 4π × 10^{–7} T m A^{–1}

| B | = 4π x 10^{-7} x 2π x 100 x 0.4 / 4π x 0.08

= 3.14 x 10^{-4} T

Hence, the magnitude of the magnetic field is 3.14 × 10^{–4} T.

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Ritesh
2019-12-03 13:57:35

Thank you bhut hard...ðð

Dipu Dileep
2019-10-31 06:15:19

Thanks ððððâð¼âð¼âð¼

Kavya
2019-07-20 19:28:33

Thank you

Saima
2019-03-01 20:03:44

Thankyou

Ankit kumar
2018-08-03 11:54:54

Question is incomplete because the value of mean radius is not given.but anyway you use the formula to find magnetic field at the center of coil that B= munot NI divided by 2 multiply mean radius

Swapna
2018-06-07 08:11:34

Can u plz post the solution for this question that a circular coil has 35 turns and a mean radius of 35 turns carries a current of 1.2 A. Find magnetic field at center of coil.

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