Question

In a solenoid of length 5cm the total number of turns is 1000 if 2a circular flow's through it turns calculate the magnetic field at a point on it axis out at the end of the solenoid

Answer 1

Answer:

$\boxed{\mathfrak{Magnetic \ field \ (B) = 8\pi \: mT}}$

Given:

Length of solenoid (l) = 5 cm = 0.05 m

Total number of turns in solenoid (N) = 1000

Current flowing through turns of solenoid (I) = 2 A

Explanation:

At a point on it's axis out at the end of the solenoid, the magnetic field is found to be:

$\boxed{ \bold{B = \dfrac{\mu_0Ni}{2l}}}$

$\rm \mu_0 \longrightarrow$ $\rm Permeability \ of \ free \ space$ ($\rm 4\pi \times 10^{-7}$)

By substituting values in the formula we get:

$\rm \implies B = \dfrac{4\pi \times {10}^{ - 7} \times 1000 \times \cancel{2}}{ \cancel{2} \times 0.05} \\ \\ \rm \implies B = \dfrac{4\pi \times {10}^{ - 7} \times {10}^{3} }{5 \times {10}^{ - 2} } \\ \\ \rm \implies B = 0.8\pi \times {10}^{ - 7 + 3 + 2} \\ \\ \rm \implies B = 0.8\pi \times {10}^{ - 2} \: T \\ \\ \rm \implies B = 8\pi \times {10}^{ - 3} \: T \\ \\ \rm \implies B = 8\pi \: mT$