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Two ideal solenoids are \"nested\" (aligned concentrically) along the same axis,

ID: 2258249 • Letter: T

Question


Two ideal solenoids are "nested" (aligned concentrically) along the same axis, as show above. The outer solenoid carries an inital current I(1) in its coils, as shown in its wires above. The inner solenoid's current, I(2), is initially zero. Assume that both solenoids have some small, but non-zero, resistance.


Outer Solenoid: L(1) = 75 cm, r(1) = 5.0 cm, N(1) = 1200 turns

Inner Solenoid: L(2) = 33 cm, r(2) = 4.0 cm, N(2) = 450 turns


Suppose you linearly increase I(1) from 5.0 A to 12.0 A during a time period of 0.33 s.


a. What is the magnetic field strength and direction inside the larger solenoid when I(1) = 5.0 A? Then when I(2) = 12.0 A? (Please show your work)


b. Find the strength and direction of the total emf induced in the inner solenoid. Draw "into page" or "out of page" current symbols on the wires of the inner solenoid to indicate the direction of the induced curret (assuming that current is allowed to flow). Please show work

Two ideal solenoids are "nested" (aligned concentrically) along the same axis, as show above. The outer solenoid carries an inital current I(1) in its coils, as shown in its wires above. The inner solenoid's current, I(2), is initially zero. Assume that both solenoids have some small, but non-zero, resistance. Outer Solenoid: L(1) = 75 cm, r(1) = 5.0 cm, N(1) = 1200 turns Inner Solenoid: L(2) = 33 cm, r(2) = 4.0 cm, N(2) = 450 turns Suppose you linearly increase I(1) from 5.0 A to 12.0 A during a time period of 0.33 s. What is the magnetic field strength and direction inside the larger solenoid when I(1) = 5.0 A? Then when I(2) = 12.0 A? Find the strength and direction of the total emf induced in the inner solenoid. Draw "into page" or "out of page" current symbols on the wires of the inner solenoid to indicate the direction of the induced curret (assuming that current is allowed to flow).

Explanation / Answer

N1 = 1200
L1 = 0.75 m
n1 = 1200/0.75 = 1600 turns/m

B = mue*n1*I1

a) I1 = 5 A

B1 = mue*n1*I1 = 1.005*10^-2 T

B2 = mue*n1*I2 = 2.412*10^-2 T

b)
N2 = 450
r2 = 0.04 cm
A2 = 3.14*0.04^2
t = 0.33 s

induced emf = N2*A2*(B2-B1)/t = 9.64 volts

the induced current direction in inner solenoid opposite to the current in outer solenoid.

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