Question

Question 4.1: 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?

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-169

Answer 1

$\frac{nui}{2r} = \frac{100 \times 0.4 \times 4\pi \times {10}^{ - 7} }{2 \times 0.08} = \pi \times {10}^{ - 4}$
B=π×10^(-4) TESLA

Answer 2

The magnetic field at the centre of a circular coil having N turns is given by
$\large{\mathbb{B}=\left(\begin{array}{c}\frac{\mu_0I}{2r}\end{array}\right)N}$

Here r is the radius of coil and I is the current passing through coil.

now,I = 0.4A , r = 8cm = 0.08 m and $\mu_0=4\pi\times10^{-7}$
so, B = (4π × 10^-7 × 0.4)/(2 × 0.08) × 100
= 3.14 × 10^-4 T

hence , magnetic field , B = 3.14 × 10^-4 T