A glass tube 1 m long and 2cm diameter is completely and uniformly wound with 900 turns of copper wire 0.9 mm diameter insulated to 1mm diameter. Find the strength of the magnetic field at the centre of the solenoid thus formed when a P.D. of 4 volts is applied to its terminals.
ρ for copper = 1.73 x 10-8 Ω
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A long wire wrapped around in a coil of loops of wire is called a solenoid. To calculate the magnetic field in a solenoid one needs to apply Ampere's Law which provides this equation:
B = UIN where U is the permeability, I is the current in the wire and N is the number of coils (or "turns of wire").
U is a constant equal to 4 Pi X10^-7 Tm/A
N is 900 for this problem.
I needs to be calculated using Ohm's Law
I = V/R
R is not given directly, however we can calculate the length of the wire by multiplying the circuference of one loop by the number of loops. In meters, the length will be Pi X D X N = Pi X 0.02 m X900 = 56.5 meters.
The resistance of the wire can be determined by multiplying the length of the wire by the linear resistance which in meters is 0.172 Ohms/meter.
Therefore R = 0.172 Ohms/m X 56.5 m = 9.73 Ohms
I = 4.0 V/9.73 Ohms = 0.411 Amps
This gives us the information to calculate the magnetic field strength inside the solenoid:
B = 4Pi X 10^-7T m/A X 0.411A X 900 turns/m = 4.65 X 10^-4 T
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