A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni

Question

2 years ago

11

tude of the magnetic field inside the solenoid.

1 answer:

irinina [24] · Answer 1 · 2023-03-25T03:50:33+03:00

The magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .

The given parameters;

The magnitude of the magnetic field inside the solenoid is calculated as;

$B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\$

where;

$\mu_o$ is the permeability of frees space = 4π x 10⁻⁷ T.m/A

$B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T$

Thus, the magnitude of the magnetic field inside the solenoid is $6.46 \times 10^{-3} \ T$ .