A solenoid with 1200 turns per meter has a diameter of 3.00 cm. A current I = 2.
ID: 1289761 • Letter: A
Question
A solenoid with 1200 turns per meter has a diameter of 3.00 cm. A current I = 2.42 A flows in the clockwise direction (when viewed from location P) in the solenoid. A rectangular loop of length L = 16.0 cm, width w = 12.5 cm, and 2 turns is centered on the axis of the solenoid.
(a) Find the magnitude of the magnetic flux through one turn of the rectangular loop. [in Wb]
(b) When the current is increased to 5.55 A, the magnitude of the induced emf in the rectangular loop is116 mV. How long did it take for the current to get to this value? [in ms]
Explanation / Answer
Mangetic field due to solenoid is
B=uo*n*I =(4pi*10-7)*2*2.42
B=6.08*10-6T
Area of rectangular loop
A=L*W=0.16*0.125=0.02 m2
Magnetic flux
phi=NAB=1200*0.02*6.08*10-6
phi=1.46*10-4 Wb
b)
Magnetic field
B2=uo*n*I2=(4pi*10-7)*2*5.55 =1.395*10-5 T
Induced emf
E=NA(dB/dt)
0.116=1200*0.02*(1.395*10-5-6.08*10-6/dt)
dt=1.63*10-3 s or 1.63 ms
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