Question

A solenoidal coil with 23 turns of wire is wound tightly around another coil with 320 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.30 cm . At a certain time, the current in the inner solenoid is 0.100 A and is increasing at a rate of 1700 A/s

Calculate the emf induced in the outer solenoid by the changing current in the inner solenoid

Answer 1

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Answer 2

Given:

• Turns of outer solenoid, $N_2=23$

• Turns of inner solenoid, $N_1=320$

• Length of inner solenoid, $l=25.0\text{cm}=0.25\text{m}$

• Diameter of inner solenoid, $d=2.30\text{cm}\Rightarrow r=1.15\times 10^{-2}\text{m}$

• Current in inner solenoid, $I=0.100\text{A}$

• Rate of current change, $\frac{dI}{dt}=1700\text{A/s}$

Step 1: Calculate mutual inductance ($M$)
The mutual inductance between two tightly coupled solenoids is:

$$M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$$

where , and $A=\pi r^2$.

$$A=\pi (1.15\times 10^{-2}{)}^2=4.154\times 10^{-4}{\text{m}}^2$$$$M=\frac{(4\pi \times 10^{-7})(320)(23)(4.154\times 10^{-4})}{0.25}$$$$M\approx 1.53\times 10^{-5}\text{H}(\text{15.3 μH})$$

Step 2: Calculate induced emf in outer solenoid
Faraday's Law:

$$\mathcal{E}=-M\frac{dI}{dt}$$$$\mathcal{E}=-(1.53\times 10^{-5})(1700)=-0.0260\text{V}$$

(Magnitude: $0.0260\text{V}$, sign indicates direction.)

Answer:
The induced emf in the outer solenoid is $\boxed{{\displaystyle 0.0260\text{V}}}$