Question

A circular loop of radius 9.7 cm carries a current 2.3 A. Obtain the magnitude of the magnetic field
(a) at the centre of the loop
(b) at a distance of 9.7 cm from the centre of the loop but on the axis.

Answer 1

Data: R = z = 9.7 cm = 9.7 x 10^-2 m, I = 2.3 A, N = l

(a) At the centre of the coil:

The magnitude of the magnetic induction,

B = `(mu_0"NI")/(2"R")`

`= ((4pi xx 10^-7)(1)(2.3))/(2(9.7 xx 10^-2)) = (2 xx 3.142 xx 2.3)/9.7 xx 10^-5`

`= 1.49 xx 10^-5` T

(b) On the axis, at a distance z = 02 m from the coil:

B = `mu_0/(4pi) (2pi"IR"^2)/(("R"^2 + "z"^2))^(3/2)`

`("R"^2 + "z"^2)^(3/2) = (2"R"^2)^(3/2) = 2sqrt2"R"^3`     (∵ R = z)

∴ B = `mu_0/(4pi) (2pi"IR"^2)/(2sqrt2"R"^3) = mu_0/(4pi) (pi"I")/(sqrt2"R")`

`= (10^-7) (3.142 xx 2.3)/(1.414 xx 9.7 xx 10^-2)`

`= 7.227/13.72 xx 10^-5 = 5.267 xx 10^-6 "T" = 5.267 mu "T"`

Answer 2

Given:

R = 9.7 cm = 9.7 × 10^-2 m, 

I = 2.3 A,

z = 9.7 cm = 9.7 × 10^-2 m

To find:

Magnetic field

1. at the centre of the loop
2. on the axis at a distance

Formulae:

1. B_c = `(mu_0"I")/(2"R")`
2. B_a = `(mu_0"I""R"^2)/(2("z"^2 + "R"^2)^{3/2})`

Calculation: 

From formula (i),

B_c = `(4pi xx 10^-7 xx 2.3)/(2 xx 9.7 xx 10^-2)`

= `(2 xx 3.142 xx 2.3)/9.7 xx 10^{-7 + 2}`

= {antilog [log 2 + log 3.142 + log 2.3 - log 9.7]} × 10^-5 

= {antilog (0.3010 + 0.4972 + 0.3617 - 0.9868)} × 10^-5 

= {antilog (0.1731)} × 10^-5 

= 1.489 × 10^-5 

≈ 1.49 × 10^-5  T

= 14.9 μT

From formula (ii),

B_a = `(4pi xx 10^-7 xx 2.3 xx (9.7 xx 10^-2)^2)/(2[(9.7 xx 10^-2)^2 + (9.7 xx 10^-2)^2]^{3/2})`

= `(4pi xx 10^-7 xx 2.3 xx (9.7 xx 10^-2)^2)/(2 xx 2^{3/2} xx (9.7 xx 10^-2)^3)`

= `(pi xx 10^-7 xx 2.3)/(2^{1/2} xx (9.7 xx 10^-2))`

= `5.268 xx 10^-6`T

= 5.268 μT

1. The magnetic field at the centre of 14.9 μT.
2. The magnetic field on the axis at a distance of 9.7 cm from the centre is 5.268 μT.