Question

A long solenoid of length 40 cm having 240 turns carries a current of 5 A. The magnetic field at the centre of the solenoid is​

Answer 1

Given:

A long solenoid of length 40 cm having 240 turns carries a current of 5 A.

To find:

Magnetic field at centre of solenoid?

Calculation:

The general expression for magnetic field intensity at centre of a solenoid is :

$\boxed{ \rm \: B = \dfrac{ \mu_{0}(i \times n) }{L} }$

• 'i' is current, 'n' is number of coil and 'L' is length of coil.

$\rm \implies\: B = \dfrac{ \mu_{0} \times 5 \times 240 }{ (\frac{40}{100} )}$

$\rm \implies\: B = \dfrac{(4\pi \times {10}^{ - 7} )\times 5 \times 240 }{ (\frac{40}{100} )}$

$\rm \implies\: B = \pi \times {10}^{ - 8} \times 5 \times 240$

$\rm \implies\: B = 1200\pi \times {10}^{ - 8}$

$\rm \implies\: B = 12\pi \times {10}^{ - 6}$

$\rm \implies\: B = 37.69 \times {10}^{ - 6} \: W/ {m}^{2}$

So, magnetic field intensity is 37.69 × 10^(-6) W/m².