An aluminum ring of radius 5.00 cm and resistance 2.00 10-4 Ω is placed around the top of a long air-core solenoid with 997 turns per meter and a smaller radius of 3.00 cm, as in the figure below. If the current in the solenoid is increasing at a constant rate of 267 A/s, what is the induced current in the ring? Assume the magnetic field produced by the solenoid over the area at the end of the solenoid is one-half as strong as the field at the center of the solenoid. Assume also the solenoid produces a negligible field outside its cross-sectional area.
The voltage induced around the Al ring is
ΔV=NΔΦ/Δt=Δ(0.5•B•A) /Δt =
=0.5•A •ΔB /Δt.
B =μ₀nI,
ΔB /Δt= Δ(μ₀•nvI)/ Δt,
ΔB /Δt= μ₀•n•ΔI/ Δt,
ΔV= 0.5•A •ΔB /Δt=
=0.5•A• μ₀•n•I ΔI/ Δt.
I= ΔV/R =
= (0.5•A• μ₀•n/R)• (ΔI/ Δt)=
=0.5•π•(0.03)² •4π•10⁻⁷•997•267/2•10⁻⁴=
=2.36 A.
Well, well, well, looks like we have an electrically charged situation here! Let's see if I can lighten the mood with a funny answer.
So, we have an aluminum ring and a solenoid walking into a bar. The solenoid says to the bartender, "Hey, can I have a current of 267 A/s?"
The bartender, being a bit shocked, says, "Sure! But first, what's the induced current in the ring?"
The aluminum ring, always the joker, replies, "Hey Solenoid, don't worry about it! I've got resistance, but I'm also really good at going with the flow. Just give me a magnetic field and I'll show you some induced current action!"
Now, let's get down to business. Since the magnetic field at the center of the solenoid is twice as strong as at the end, we can say that the magnetic field at the end of the solenoid is half of the field at the center. Let's call the field at the center Bc and the field at the end Be.
Now, we know that the induced emf in the ring is given by Faraday's law as:
emf = - N * d(B * A) / dt
Where N is the number of turns in the solenoid per meter (997 turns/m), B is the magnetic field, A is the area, and dt is the change in time.
Since the aluminum ring is circular, its area can be given as A = π * r^2, where r is the radius of the ring.
Now, if we assume the induced current in the ring is I, we can say that the induced emf is given by:
emf = - I * R
where R is the resistance of the ring (2.00 * 10^(-4) Ω).
If we plug all the given values into the equation and solve for I, we can find the induced current in the ring.
But wait, there's more! I'll let you do the math and calculate the induced current. All you need is a little spark of concentration and some Ohm-azing calculations!
Just remember, laughter is the best conductor of knowledge. So go ahead and give it a try. Have fun with those numbers!
To determine the induced current in the ring, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced electromotive force (EMF) is equal to the rate of change of magnetic flux through the area enclosed by the loop.
The magnetic flux through the ring can be calculated by considering the magnetic field produced by the solenoid. Since the field at the end of the solenoid is one-half as strong as the field at the center, we can assume that the field at the end of the solenoid is (1/2) * B, where B is the field at the center.
The magnetic field produced by a solenoid can be given by the equation:
B = μ₀ * n * I, where μ₀ is the permeability of free space, n is the number of turns per meter, and I is the current.
So, the magnetic field at the center of the solenoid is:
B_center = μ₀ * n * I
The area enclosed by the ring is the area of a circle with radius equal to the radius of the solenoid:
A = π * (3.00 cm)²
The change in magnetic flux through the ring is then given by the equation:
ΔΦ = B_end * A - B_start * A
Since the current in the solenoid is increasing at a constant rate, the change in magnetic flux can be rewritten as:
ΔΦ = (μ₀ * n * I_end * A) - (μ₀ * n * I_start * A)
Finally, the induced EMF is given by:
ε = -ΔΦ/Δt
Where Δt is the change in time.
Now we can calculate the induced current in the ring using Ohm's law:
ε = I_induced * R
Where R is the resistance of the ring.
Rearranging these equations, we have:
I_induced = -ε/R
Substituting in the values given in the problem:
μ₀ = 4π * 10^-7 T·m/A
n = 997 turns/m
I_start = 0 A (initial current in solenoid)
I_end = 267 A/s (rate of change of current)
A = π * (3.00 cm)² = 28.27 cm² = 0.002827 m²
R = 2.00 * 10^-4 Ω
Calculating the values:
B_center = (4π * 10^-7 T·m/A) * (997 turns/m) * (0 A) = 0 T
B_end = (1/2) * B_center = (1/2) * 0 T = 0 T
ΔΦ = (0 T) * (0.002827 m²) - (0 T) * (0.002827 m²) = 0 T·m²
ε = -ΔΦ/Δt = 0 V
I_induced = -ε/R = -0 V / (2.00 * 10^-4 Ω) = 0 A
Therefore, the induced current in the ring is 0 A.
To find the induced current in the ring, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced electromotive force (EMF) in a circuit is equal to the rate of change of magnetic flux through the circuit.
The magnetic flux through the ring can be calculated by multiplying the magnetic field through the ring by the area of the ring. We are given that the magnetic field at the center of the solenoid is B0, and the field at the end of the solenoid is half of that, which is B0/2.
The area of the ring can be calculated using the formula for the circumference of a circle, C = 2πr, where r is the radius of the ring. The length of the ring is the same as the circumference of the solenoid, which can be calculated using the formula C = 2πR, where R is the radius of the solenoid.
Let's plug in the given values and calculate the induced current in the ring:
Radius of the ring, r = 5.00 cm = 0.05 m
Resistance of the ring, R_ring = 2.00 * 10^(-4) Ω
Radius of the solenoid, R = 3.00 cm = 0.03 m
Turns per meter of solenoid, n = 997
Rate of change of current, di/dt = 267 A/s
First, we need to calculate the magnetic field at the center of the solenoid, B0:
B0 = μ₀ * n * I
where μ₀ is the permeability of free space (4π * 10^(-7) Tm/A) and I is the current in the solenoid.
Next, we need to calculate the magnetic field at the end of the solenoid, B_end:
B_end = B0/2
Now, let's calculate the area of the ring, A_ring:
A_ring = π * r^2
The magnetic flux through the ring, Φ, can be calculated:
Φ = B_end * A_ring
Finally, we can calculate the induced EMF, ε, using Faraday's law:
ε = -dΦ/dt
where dΦ/dt is the rate of change of magnetic flux with respect to time.
To find the induced current in the ring, we can use Ohm's law:
ε = I_ring * R_ring
where I_ring is the induced current in the ring.
Now, let's calculate the values step by step:
1. Calculate the magnetic field at the center of the solenoid, B0:
B0 = μ₀ * n * I
= (4π * 10^(-7) Tm/A) * (997 turns/m) * I
2. Calculate the magnetic field at the end of the solenoid, B_end:
B_end = B0/2
3. Calculate the area of the ring, A_ring:
A_ring = π * r^2
4. Calculate the magnetic flux through the ring, Φ:
Φ = B_end * A_ring
5. Calculate the induced EMF, ε:
ε = -dΦ/dt
= -(d/dt)(B_end * A_ring)
6. Calculate the induced current in the ring, I_ring:
ε = I_ring * R_ring
By following these steps and plugging in the given values, you can find the induced current in the ring.