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Question

https://session.masteringphysics.com/myct/itemView?assignmentProblemID=56219423

The figure shows two 28-turn coils tightly wrapped on the same 2.0-cm-diameter cylinder with 1.0-mm-diameter wire. The current through coil 1 is shown in the graph. A positive current is into the page at the top of a loop. Assume that the magnetic field of coil 1 passes entirely through coil 2.

it is not allowing me to add the picture but it has a graph that has no slope from 0-.1 and .3 to .4 and a positive slope of 20 from .1 to .3

Explanation / Answer

use the formula for Self inductance of coil 1, L1 = u*N^2*A/l

where A = area of the loop = pi*R^2 = pi*0.01^2 = 3.14*10^-4 m2

and , uo is 4pi e 7 = 1.26*10-6

l = length of the loop = 28*d = 28*0.0011 = 0.0308 m

So, inductance L1 = 1.26*10^-6*28^2*3.14*10^-4/(0.0308) = 1*10^-5 H

So, flux linked in the coil 2, Q = N L i

Due to change in current, induced emf = N*L1*d(I1)/dt

where d(I1)/dt = rate of change of current I1 which gives the slope of the curve

at t = 0.25 s, dI1/dt = 4/0.2 = 20 A/s

so, Induced emf = 28*1*10^-5*20 = 0.0056 V

So, Induced current, 0.0056/2 = 0.0028 A = 2.8 mA

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