Question

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 magnitude of the magnetic field inside the solenoid.

Answer 1

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

The given parameters;

• length of the solenoid, L = 91 cm = 0.91 m
• radius of the solenoid, r = 1.5 cm = 0.015 m
• number of turns of the solenoid, N = 1300
• current in the solenoid, I = 3.6 A

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 $6.46 \times 10^{-3} \ T$.

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