Question

In the figure below a thin coil of radius r1= 4 cm, containing N1= 5300 turns, is connected through a resistor R = 100 Ω to an AC power supply running at a frequency f= 2100 so that the current through the resistor (and coil) is I1sin(2π×2100t). The voltage across the resistor triggers an oscilloscope that also displays this voltage, which is 10 V peak-to-peak and therefore has an amplitude of 5 V (and therefore the amplitude of the current is I1= 0.05 A).



A second thin coil of radius r2= 2 cm, containing N2= 3100 turns, is a distance L= 40 cm from the first coil. The axes of the two coils are along the same line. The second coil is connected to the upper input of the oscilloscope, so that the voltage across the second coil can be displayed along with the voltage across the resistor.

Calculate the amplitude (maximum voltage) of the second coil voltage.

V

If you must make simplifying assumptions, state clearly what they are.

Our assumptions are that L≪r1 and that the coils are thick coils.

Our assumptions are that L≫r1 and that the coils are thin coils.

Answer 1

ma Magnetic field B, due to coil of N, to radius & distance L, current I is Biz Ni Mo (21 ri - 47 (2)3 &z N₂ B, ucre which is flux through second coil of radius ele having N2 turns. 02 Na N, MO (21 4,5 (2) Ezt HT ILI3 ET-M, Ng (20 MIL QT X ) SIO IzI sin (2rrt) I a Il sin (20 (21000)) — (given) 1.os sin (4200 Lt) Cil, = 0.05) d] =(0.05) (4200 J) cos (42003 ) at I Z

- 4 cm 0.04 m. N = 5300, R-1002, =2100 I = I sin (21T X2 10ot), I1=0.05A, dlz=2 an 2002m. L=40cm= 0.40m, N2 = 3100 Putting all values in (3) E = - (5300)(3100) (47X16) 2(3014)*(004%%-02) 4T (0.40)3 (0.05) (42001) cos (42001Tt) voltage will be maximum when cos (42005t) = 1 E =-(5300) (3100) (41) X16) (2) (3.14)?(0.04)? (0.022 (0.05) ( 4200 s) 4 TT (0.400)3 TE= 0.214 v Assumptions made are & L77r, and coils. are thin coils