Question

A long solenoid has a length of 0.67 m and contains 1700 turns of wire. There is a current of 5.5 A in the wire. What is the magnitude of the magnetic field within the solenoid?

Answer 1

B = 17.5mT.

A solenoid is a coil formed by a wire (usually copper) wound into a cylindrical spiral shape capable of creating a magnetic field that is extremely uniform and intense inside, and very weak outside.

To calculate the magnetic field generated inside the solenoid through which a current flows is given by the equation:

B = μ₀nI

Where  μ₀ is the constant of magnetic proportionality of the vacuum (4π x 10⁻⁷T.m/A), n is the relation between the number of turns of wire and its length given by N/L and I is the current flowing through the solenoid.

Given a long solenoid of length 0.67m, 1700.00 turns of wire and a current flowing through the wire of 5.50A. Calculate the magnetic field inside the solenoid.

B = (4π x 10⁻⁷T.m/A)(1700turns/0.67m)(5.50A)

B = 0.0175T

B = 17.5mT