In the long solenoid shown in the figure, the number of turns per unit of length
ID: 1411438 • Letter: I
Question
In the long solenoid shown in the figure, the number of turns per unit of length is n_1 = 210 turns/cm. At the center of this solenoid there is a shorter solenoid made of N_2 = 150turns closely spaced. The diameter of the smaller solenoid is d_2 = 2.3cm, and its axis coincides with the axis of the longer solenoid. The current in the first solenoid grows linearly by an amount Delta i = 2 A in a time of Delta t = 0.19 s. Calculate the rate of change of the current; Delta i/Delta t. Calculate the rate of change of the magnetic field at the center of the long solenoid. Calculate the rate of change of the magnetic field flux in one of the turns of the smaller solenoid. Calculate the absolute value of the e. m. f. induced in the smaller solenoid while the current in the longer solenoid is increasing. Remember: mu_0 = 4pi times 10^-7 T m/AExplanation / Answer
(a) ans
the rate of change ofthe current =i/t=2/0.19 =10.53 A/s
(b) ans
number turns of centre of long solenoid n1=210 turns/cm
current i=2A
time t=0.19 s
the magnetic field at the centre of solenoid B=µ n1 i=4*3.14*10^-7*210*10^2*2=0.052752 T
the rate of change of magnetic field =B/t=0.052752/0.19=0.28 T/s
(c) ans
the magnetic flux =µ n1N2(3.14*d2^2) i
=4*3.14*10^-7*210*10^2 *2*150*3.14*(2.3*10^-2)^2
=1.3*10^-2 weber
the rate of change of magnetic fiux =((magnetic fiux/time)
=1.3*10^-2/0.19
=6.84*10^-2 wb/sec
(d) ans
first we find the mutual inductance
therefore the mutual inductance =>M=µ n1 N2 A
=4*3.14*10^-7*210*10^2*150*3.14*(2.3*10^-2)^2
=6.572*10^-3
the induced e.m.f e=M di/dt
=6.572*10^-3*10.53
=69.2*10^-3 V
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