Question 16: A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A . The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4 \pi \times 10^{-7} SI units) :
(1) 44 mT
(2) 4.4 T
(3) 4.4 mT
(4) 44 T
Answer: Option (3)
Explanation:
For a circular coil having N turns, radius R,
and carrying current I, the magnetic field at the centre is given by
B=\frac{\mu_{0} N I}{2 R}Here, N=100, I=7 \ \mathrm{A},
and the radius R=10 \ \mathrm{cm}=0.1 \ \mathrm{m}.
The permeability of free space is \mu_{0}=4 \pi \times 10^{-7} SI units.
Substituting the values,
B=\frac{(4 \pi \times 10^{-7}) \times 100 \times 7}{2 \times 0.1} B=\frac{2800 \pi \times 10^{-7}}{0.2} B=14000 \pi \times 10^{-7} \ \mathrm{T}Using \pi \approx 3.14,
B \approx 4.4 \times 10^{-3} \ \mathrm{T} B = 4.4 \ \mathrm{mT}Hence, the correct answer is Option (3).