Answer:
The magnitude of the magnetic field near the center of the solenoid is 11.3 mT
Explanation:
No. of turns [tex]N = 300[/tex]
Radius [tex]r = 0.040[/tex] m
Length of solenoid [tex]L= 40 \times 10^{-2}[/tex] m
Current [tex]I = 12[/tex] A
For finding the magnetic field near the center of the solenoid is given by,
[tex]B = \frac{\mu _{o} NI }{L}[/tex]
Where [tex]\mu _{o} = 4\pi \times 10^{-7}[/tex]
[tex]B = \frac{4 \pi \times 10^{-7} \times 300 \times 12}{40 \times 10^{-2} }[/tex]
[tex]B = 11.3 \times 10^{-3}[/tex] T
[tex]B = 11.3[/tex] mT
Therefore, the magnitude of the magnetic field near the center of the solenoid is 11.3 mT
