Question

A circular coil of wire consisting of 100 turns each of radius 8 cm carries a current of 0.40 a what is the magnitude of magnetic field b at the centre of the coil

Answer 1

Answer:

B=3.14 * 10^(-5) Tesla

Explanation:

Given N = 100. I=0.4 A

Radius r = 8 cm = 0.8m

We have formula

B=( M• *N*I)/2r

Where M• = 4π *10^-7 Am

Now

B=4π*10^(-7) * 100*0.4/(2*0.8)

On solving we get

B= π*10^(-5)

=3.14 * 10^(-5)

Answer 2

Explanation:

Number of turns on the circular coil, n = 100

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

Current flowing in the coil, I = 0.4A

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

|B| = $\sf{\dfrac{\mu_0}{4π}}$$\sf{\dfrac{2π\:nl}{r}}$

Where,

$\sf{\mu_0}$ = Permeability of free space

= $\sf{4π\:x\:{10^{-7}}}$ $\sf{Tm}$$\sf{A^{-1}}$

|B| = $\sf{4π\:x\:{10^{-7}}}$/4π * $\sf{2π\:x\:100\:x\:0.04}$/0.08

= $\sf{3.14\:x\:{10^{-4}}}$$\sf{T^{-1}}$