Question: 8: A closely packed coil having 1000 turns has an average radius of 62.8 cm . If current carried by the wire of the coil is 1 A , the value of magnetic field produced at the centre of the coil will be (permeability of free space =4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m} ) nearly:
(1) 10^{-1} \mathrm{~T}
(2) 10^{-2} \mathrm{~T}
(3) 10^{2} \mathrm{~T}
(4) 10^{-3} \mathrm{~T}
Answer: Option (4)
Explanation:
The magnetic field at the centre of a circular coil having N turns, radius R, and carrying current I is given by:
B=\frac{\mu_0 N I}{2R}Given, number of turns N=1000, current I=1\ \mathrm{A},
radius R=62.8\ \mathrm{cm}=0.628\ \mathrm{m},
and permeability of free space \mu_0=4\pi\times10^{-7}.
Substituting the values:
B=\frac{4\pi\times10^{-7}\times1000\times1}{2\times0.628} B=\frac{4\pi\times10^{-4}}{1.256} B\approx10^{-3}\ \mathrm{T}Hence, the magnetic field produced at the centre of the coil is approximately
10^{-3}\ \mathrm{T}.