1) A circular solenoid contains 100 turns of wire and carries a current of 2 amp
ID: 2120504 • Letter: 1
Question
1) A circular solenoid contains 100 turns of wire and carries a current of 2 amps. Its radius is 1 cm and its length is 10 cm. Calculate the magnetic flux through one turn of the wire making the solenoid.
%u03A6B =
2)
A rectangular loop rotates with a constant angular velocity about an axis as shown. It makes one complete rotation in 0.04 s. The magnetic field lines are straight and parallel as shown. They are perpendicular to the axis of rotation. The lengths are c= 0.8 m and d= 4.3 m, the strength of the magnetic field is 0.08 T. The total resistance of the loop is 160 %u03A9.
A)
F(0o) =
B)
F(30o) =
C)
F(90o) =
f =
E)
I =
F)
t =
Explanation / Answer
(1) n = N/L = turns per unit length
B = u*(n)*I = 4pi*10^-7*(100/0.1)*2 = 8*pi*10^-4 T
Flux = B*S = 8*pi*10^-4*pi*(10^-2)^2 = 8*pi^2*10^-8 Wb = 7.88*10^-7 Wb
(2)
Angular velocity =w = 2*pi/T = 2*pi/0.04 = 50*pi
B = 0.08 T
R = 160 Ohm
Area = d*2c = 4.3*1.6 = 6.88 m^2
(A) Flux = B.A = B*A*cos(u)
Where u is the angle between normal and magnetic field.
Flux = B*A*cos(0) = 0.08*6.88*1 = 0.5504 T-m^2
(B) Flux = B*A*cos(30) = 0.08*6.88*0.866 = 0.476 T-m^2
(C) Flux = B*A*cos(90) = B*A*0 = 0 T-m^2
(D) Emf = -d(FLUX)/dt = B*A*sin(u)*du/dt = B*A*w*sin(u)
Only sin(u) is variable which is max at u = 90 degrees
emf in the loop is at its maximum value at 90 degrees
(E) Max I = Max Emf/R = B*A*w/R = 0.08*6.88*50*pi / 160 = 0.54 Amp
(F) Torque = [2*I*(L x B)]*r
L x B = L*B*sin(90) = L*B = 4.3*0.08 = 0.344
Where I Is current Max
L IS THE length of vetical wires = d = 4.3 m
r is radius of rotation of vertical wires = 0.8 m
Torque = 2*0.54*0.344*0.8 = 0.297 N-m
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