It is desired to construct a solenoid that will have a resistance of 5.65 Ohm (a

Question

It is desired to construct a solenoid that will have a resistance of 5.65 Ohm (at 20 degree C) and produce a magnetic field of 4.00 x 10^-2 T at its center when it carries a current of 3.75 A. The solenoid is to be constructed from copper wire having a diameter of 0.500 mm. If the radius of the solenoid is to be 1.00 cm, determine the following. (The resistivity of copper at 20 degree C is 1.7 x 10^-8 Ohm. m.) the number of turns of wire needed to build the solenoid the length the solenoid should have

Explanation / Answer

here,
resistance in solenoid wire, R = 5.65 ohms
Current , I = 3.75 A
diameter of solenoid wire, ds = 0.5 mm = 0.0005 m
magnatic field , B = 4*10^-2 T
Resistivity of copper wire, p = 1.7 * 10^-8 ohm.m

radius of solenoid, rs = 1cm = 0.01 m

Magnatic field in solenoid is given as:
B = uo * (N/L) * I -------------------------(1)

Part 1 :
resistance in wire is given as,
R = p*L/A

where, A is area of wire, L is length
L = R*A/p
L = (5.65 * pi * (0.0005/2)^2)/(1.7 * 10^-8)
L = 65.257 m

Each turn of the solenoid is pi*d = *2*0.01 = 0.063 m
so there are 65.257/0.063 = 1036 turns

Part 2:
Equation 1 can be written as :

L = (uo*N*I)/B
L = (4*3.14*10^-7 * 3.75 * 1036)/(4*10^-2)
L = 0.122 m

Length of solenoid is 12.2 cm

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