Question

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 ×102 T, what is the number of turns per meter for this solenoid?

Answer 1

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Step-by-step explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

`B =\mu_o I((N)/(L) )\\\\B = \mu_o I(n)\\\\n = (B)/(\mu_o I) \\\\n = (2.8 * 10^(-2))/(4 \pi * 10^(-7) * 5.0) \\\\n = 4.5 * 10^3 \ turns/m`

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.