Answer:
The magnitude of the magnetic field inside the solenoid is 1.5 × 10⁻²T
Explanation:
Given that,
Length = 40 cm
Radius = 0.040 cm
Number of turns = 400
Current = 12 A
We need to calculate the magnetic field
Using formula of magnetic field inside the solenoid
[tex]B =\mu nIB =\mu\dfrac{N}{l}I[/tex]
Where,[tex]\dfrac{N}{l}[/tex]=Number of turns per unit length
I = current
B = magnetic field
Put the value into the formula
[tex]B =4\pi\times10^{-7}\times\dfrac{400}{40\times10^{-2}}\times12B = 1.5\times10^{-2}\ T[/tex]
Hence, The magnitude of the magnetic field inside the solenoid is 1.5 × 10⁻²T
