A solenoid of length 50 cm and radius of cross section 1 cm has 1000 turns of wire wound over it.

Question

A solenoid of length \(50 \mathrm{~cm}\) and radius of crosssection \(1 \mathrm{~cm}\) has 1000 turns of wire wound over it. If the current carried is \(5 \mathrm{~A}\), the magnetic field on its axis, near the centre of the solenoid is approximately (permeability of free space \(\mu_{0}=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}\) )

(a) \(0.63 \times 10^{-2} \mathrm{~T}\)

(b) \(1.26 \times 10^{-2} \mathrm{~T}\)

(c) \(2.51 \times 10^{-2} \mathrm{~T}\)

(d) \(6.3 \mathrm{~T}\)

Answer 1

The correct option of this question will be (b).

Solution —

The magnetic field is given by

$B=\mu_{0} n I$

where, $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}{ }^{-1}$

$ n =\frac{1000}{50 \times 10^{-2}}$ $I=5 \mathrm{~A}$

$\therefore \quad B =4 \pi \times 10^{-7} \times \frac{1000}{50 \times 10^{-2}} \times 5$

$B =1.26 \times 10^{-2} \mathrm{~T}$