Question

A 290-turn solenoid having a length of 32 cm and a diameter of 11 cm carries a current of 0.30 A. Calculate the magnitude of the magnetic field inside the solenoid.

Answer 1

The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

What's the expression of magnetic field inside a solenoid?

• Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I

• n = no. of turns per unit length and I = current through the solenoid

What's is the magnetic field inside the solenoid here?

• Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

• So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

Learn more about the magnetic field inside the solenoid here:

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