Answer:
The magnetic field at the center of the solenoid is 7.53 mT.
Explanation:
Given that,
Number of turns in the solenoid, N = 300
Length of the wire, L = 20 cm = 0.2 m
Diameter of each coil, d = 0.8 cm
Radius, r = 0.4 cm
Current in the solenoid, I = 4 A
We need to find the magnetic field at the center of solenoid. We know that the magnetic field is given by the below formula as :
[tex]B=\mu_o nI[/tex]
n is number of turns per unit length
[tex]B=\dfrac{\mu_o NI}{L}[/tex]
So,
[tex]B=\dfrac{4\pi \times 10^{-7}\times 300\times 4}{0.2}\\\\B=7.53\times 10^{-3}\ T\\\\B=7.53\ mT[/tex]
So, the magnetic field at the center of the solenoid is 7.53 mT.
