A solenoid 15cm long has 300 turns of wire. A current of 5A flows through it. What is the magnitude of the magnetic field inside the solenoid?

Correct answer: C. 4πx10-3T

  • A. 75 x 10^7T
  • B. 60 x 10^4T
  • C. 4πx10-3T
  • D. 750π x103T

Explanation

The magnetic field inside a solenoid can be calculated using the formula:B = µ0 x n x Iwhereµ0 = permittivity of free space = 4πx10-7I = Current = 5An = is the number of turns per unit length (turns per meter) = 300/0.15 = 2000Substitute these values in the formulaB = 4πx10-7 x 5 x 2000.B = 4πx10-7 x 1x104B = 4πx10-3.

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