Induced current of a coil surrounding a solenoid

In summary: It happens. Let's try going through the problem again with the correct radius.In summary, using the correct radius for the solenoid, the magnitude of the induced current in the 170-turn coil is 0.053 A. The direction of the current in the 12 ohm resistor was not specified in the conversation. To calculate the induced emf, we used the equation -(mu_0)(n)(A)(dI/dt), where n is the turns per length in the solenoid with changing current dI/dt and the area of the solenoid, A. The B field is confined to the solenoid, so the area of the solenoid must be used in the calculation.
  • #1
jchoca
4
0

Homework Statement


GIANCOLI.ch29.p67.jpg

A coil with 170 turns, a radius of 5.0 cm, and a resistance of 12 ohms surrounds a solenoid with 230 turns/cm and a radius of 4.7 cm; see the figure. The current in the solenoid changes at a constant rate from 0 to 1.8 A in 0.11 s.

Calculate the magnitude and direction of the induced current in the 170-turn coil.

Homework Equations


EQ1: induced emf (of a circular conducting loop surrounding a solenoid)
= -(mu_0)(n)(A)(dI/dt)
where n is the turns per length in the solenoid with changing current dI/dt and loop of area A.
EQ2: induced emf = -(dflux/dt)
EQ3: flux = BA
EQ4: B_solenoid = (mu_0)(n)(I)
EQ5: I_induced = (induced emf)/R

The Attempt at a Solution



I tried to take equation 1 and compute the emf for a single loop, then I multiplied that times 170. I then divided that by 12 to get the current. However, MP says it's wrong. My result was 0.053 A using n = 23,000 turns/m, A = pi * 0.05^2, and dI/dt = 1.8/0.11.

Note: Problem is #67 from Ch 29 of Physics for Scientists and Engineers with Modern Physics By Douglas C. Giancoli, 4th ed.
 
Last edited:
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  • #2
I got the same number for the current as you. What direction did you give for the current in 12 Ω resistor? Maybe that's what you got wrong.
 
  • #3
jchoca said:
EQ1: induced emf (of a circular conducting loop surrounding a solenoid)
= -(mu_0)(n)(A)(dI/dt)
where n is the turns per length in the solenoid with changing current dI/dt and loop of area A.

The B field is confined to the solenoid. Instead of the area of the loop, you should use the area of the solenoid.

jchoca said:
My result was 0.053 A using ... A = pi * 0.05^2 ...
This is not the correct radius for calculating the area of the solenoid.
 
  • #4
TSny said:
This is not the correct radius for calculating the area of the solenoid.
Oops, I missed that even though I looked to make sure that the correct solenoid radius was used. :headbang:
 
  • #5
kuruman said:
Oops, I missed that even though I looked to make sure that the correct solenoid radius was used.
No problem. We're going to miss things.
 

1. What is induced current of a coil surrounding a solenoid?

The induced current of a coil surrounding a solenoid is the electric current that is produced in the coil when there is a change in the magnetic field of the solenoid. This change in magnetic field can be caused by either the solenoid itself or an external magnetic field.

2. How is the induced current of a coil surrounding a solenoid calculated?

The induced current of a coil surrounding a solenoid can be calculated using Faraday's Law of Induction. This law states that the induced electromotive force (EMF) in a closed loop is equal to the negative rate of change of the magnetic flux through the loop. This can be represented by the equation EMF = -N(dΦ/dt), where N is the number of turns in the coil and Φ is the magnetic flux.

3. What factors affect the strength of the induced current in a coil surrounding a solenoid?

The strength of the induced current in a coil surrounding a solenoid depends on several factors, including the strength of the magnetic field, the number of turns in the coil, the rate of change of the magnetic field, and the resistance of the coil. A stronger magnetic field, more turns in the coil, and a faster rate of change will result in a stronger induced current, while a higher resistance will decrease the strength of the induced current.

4. What is the purpose of using a coil surrounding a solenoid?

The purpose of using a coil surrounding a solenoid is to increase the strength of the magnetic field and therefore, increase the induced current. The coil acts as an amplifier, allowing for a larger current to be induced in the coil compared to a single wire. This is useful in applications such as generators and transformers.

5. Can the direction of the induced current be predicted in a coil surrounding a solenoid?

Yes, the direction of the induced current in a coil surrounding a solenoid can be predicted using Lenz's Law. This law states that the direction of the induced current will always be such that it opposes the change in magnetic flux that caused it. This can be determined using the right-hand rule, where the direction of the induced current is in the opposite direction of the change in magnetic field.

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