Question

A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns of wire per centimeter, and the windings carry a current of 0.275 AS. A second coil having N turns and a larger diameter is slipped over the solenoid so that the two are coaxial. The current in the solenoid is ramped down to zero over a period of 0.50 s.
(a) What average emf is induced in the second coil if it has a diameter of 3.90 cm and N = 48? Express your answer in microvolts to two significant figures.
(b) What is the induced emf if the diameter is 7.80 cm and N = 96? Express your answer in volts to two significant figures.

Answer 1

Answer with Explanation:

We are given that

Diameter,d1=3 cm

Radius,$r_1=\frac{d_1}{2}=\frac{3}{2}=1.5 cm=\frac{1.5}{100}=0.015 m$

1 m=100 cm

Number of turns per cm,n=40

Number of turns per  m,n=4000

Current,I=0.275 A

Time period,$\Delta t$=0.5 s

a.$d'=3.9 cm$

$r'=\frac{d'}{2}=\frac{3.9}{2}=1.95 cm=0.0195 m$

Magnetic field,B=$\mu_0 nI=4\pi\times 10^{-7}\times 4000\times 0.275=1.38\times 10^{-3} T$

Magnetic flux,$\phi_1=NBA=NB(\pi r'^2)$

Final flux,$\phi_2=0$ because $I_2=0$

$E=\frac{\phi_1-\phi_1}{\Delta t}=\frac{NB(\pi r^2)-0}{0.5}$

Substitute N=48

$E=\frac{48\times 1.38\times 10^{-3}\times 3.14(0.015)^2}{0.5}=94\times 10^{-6} V=94\mu V$

b. For N=96

$E=\frac{96\times 1.38\times 10^{-3}\times 3.14(0.015)^2}{0.5}=1.9\times 10^{-4}V$