Question

Suppose you have a 34.9 m length of copper wire. If the wire is wrapped into a solenoid 0.240 m long and having a radius of 0.0510 m, how strong is the resulting magnetic field in its center when the current is 11.0 A?

Answer 1

Answer:

the strength of the magnetic field inside the solenoid is 6.278 x 10⁻³ T.

Explanation:

Given;

length of the wire, = 34.9 m

length of solenoid, L = 0.24 m

radius of the solenoid, r = 0.051 m

current in the solenoid, I = 11.0 A

The number of turns of the wire is calculated as follow;

$N = \frac{34.9}{2\pi \times 0.051} = 109 \ turns$

The strength of the magnetic field inside the solenoid is calculated as follows;

$B = \mu_0 (\frac{N}{L} )I\\\\B = 4\pi \times 10^{-7} \times (\frac{109}{0.24} )\times 11.0 \\\\B = 6.278 \times 10^{-3} \ T$

Therefore, the strength of the magnetic field inside the solenoid is 6.278 x 10⁻³ T.