A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each.

Question:

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of inside the solenoid near its centre.

Solution:

Length of the solenoid, l = 80 cm = 0.8 m

There are five layers of windings of 400 turns each on the solenoid.

$\therefore$ Total number of turns on the solenoid, $N=5 \times 400=2000$

Diameter of the solenoid, D = 1.8 cm = 0.018 m

Current carried by the solenoid, I = 8.0 A

Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,

$B=\frac{\mu_{0} N I}{l}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}$

$B=\frac{4 \pi \times 10^{-7} \times 2000 \times 8}{0.8}$

$=8 \pi \times 10^{-3}=2.512 \times 10^{-2} \mathrm{~T}$

Hence, the magnitude of the magnetic field inside the solenoid near its centre is $2.512 \times 10^{-2} \mathrm{~T}$.

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