Question

0: Due to 10 Amp of current flowing in a circular coil of \(10 \mathrm{~cm}\) radius, the mag. field produced at its centre is \(\pi \times 10^{-3}\) Tesla. The number of turns in the coil will be (a) 5000 (b) 100 (c) 50 (d) 25

Short answer

The number of turns in the coil is 50 (option c).

Step-by-step solution

Write down the given information and the formula

Given: Current, I = 10 A Radius, r = 10 cm = 0.1 m (converted to meters) Magnetic field at the center, B = \(\pi \times 10^{-3}\) T The formula for the magnetic field produced by a current-carrying circular coil at its center is: \[B = \frac{\mu_0 n I}{2r}\] where B is the magnetic field, \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{Tm/A}\)), n is the number of turns, I is the current, and r is the radius of the coil.

Solve for the number of turns (n)

We need to find the value of n. So, we will rearrange the formula for n and then substitute the given values. \[n = \frac{2rB}{\mu_0 I}\] Substituting the given values: \[n = \frac{2(0.1)(\pi \times 10^{-3})}{(4\pi \times 10^{-7})(10)}\]

Calculate the number of turns

Now, we will perform the calculation: \[n = \frac{(0.2\pi \times 10^{-3})}{(4\pi \times 10^{-7})(10)}\] \[n = \frac{0.2\pi \times 10^{-3}}{4\pi \times 10^{-6}}\] \[n = \frac{0.2}{4} \times 10^{3}\] \[n = 0.05 \times 10^{3}\] \[n = 50\] So, the number of turns in the coil is 50. The correct option is (c) 50.