Question

A solenoid with 300 turns has a radius of 0.040 m and is 40 cm long. If this solenoid carries a current of 12 A, what is the magnitude of the magnetic field near the center of the solenoid? (μ 0 = 4π × 10-7 T · m/A)

Answer 1

Answer:

The magnitude of the magnetic field near the center of the solenoid is 11.3 mT

Explanation:

No. of turns $N = 300$

Radius $r = 0.040$ m

Length of solenoid $L= 40 \times 10^{-2}$ m

Current $I = 12$ A

For finding the magnetic field near the center of the solenoid is given by,

   $B = \frac{\mu _{o} NI }{L}$

Where $\mu _{o} = 4\pi \times 10^{-7}$

   $B = \frac{4 \pi \times 10^{-7} \times 300 \times 12}{40 \times 10^{-2} }$

   $B = 11.3 \times 10^{-3}$ T

   $B = 11.3$ mT

Therefore, the magnitude of the magnetic field near the center of the solenoid is 11.3 mT