Direction and magnitude of electric field produced by current change

[songoku]

Homework Statement

Please see below

Relevant Equations

Faraday Law

The answer key is:
a) $\frac{\mu_o \pi nar_{1}^{2}}{R}$

b) $-\frac{\mu_o abn}{2}$

My questions:
1) Why is there no $\cos \theta$ in answer (a)?
$$\text{induced current}=\frac{\varepsilon}{R}$$

When calculating induced emf $\varepsilon=-\frac{d\Phi}{dt}=-\frac{d(\vec B . \vec A)}{dt}$, wouldn't the magnetic field produced by the solenoid making angle $\theta$ with respect to normal of plane of loop of wire so the rate of change of magnetic flux experienced by the loop is $\mu_o an \pi r_{1}^{2} \cos \theta$?


2) Why is there no $N$ in answer (b), where $N$ is number of turns of the solenoid?
$$\int \vec E . d\vec s=-\frac{d\Phi}{dt}$$

$\Phi$ is magnetic flux linkage so shouldn't $\Phi=N\phi$ so there is $N$ in the final answer?

3) Is the direction of electric field at distance $b$ in positive y direction?

Thanks

 

[kuruman]

My questions:
1) Why is there no cos⁡θ in answer (a)?

Because whoever wrote the answer didn't put it in?

Clearly, if θ = 90°, there will be no changing flux through the loop and hence no induced current.

Why is there no N in answer (b), where N is number of turns of the solenoid?

Because the magnetic field inside a very long solenoid of $N$ turns and length $L$ is approximated as $B=\mu_0 n I$ where $n=\dfrac{N}{L}$ is the number of turns per unit length.

 

[DaveE]

1) Why is there no cos⁡θ in answer (a)?

The induced current is the result of the magnetic flux that couples (i.e. flows through) the wire loops in the solenoid and the outer sensing coil. This is Faraday's Law. The solenoid flux in one direction is essentially entirely contained within the solenoid.

So if the sensing coil wraps around the solenoid (sort of tightly*) the angle doesn't matter, the flux through it will be the same.

An analogy is if you were counting cars on a freeway that pass through an imaginary plane, it wouldn't matter if that plane was tilted, as long as every car had to pass through it somewhere.

* You don't want that outer coil to capture much of the returning flux. A gigantic outer coil might not see any net flux, it could all (mostly) cancel.

 

[DaveE]

https://www.khanacademy.org/science...solenoids/v/magnetic-fields-through-solenoids

 

[songoku]

Because the magnetic field inside a very long solenoid of $N$ turns and length $L$ is approximated as $B=\mu_0 n I$ where $n=\dfrac{N}{L}$ is the number of turns per unit length.

I mean why for the flux it is not $\frac{\mu_o N^2 I}{L}$. I saw something similar to this when calculating self inductance of solenoid; I need to multiply the flux to get total flux linkage so the self inductance can be $\frac{\mu_o N^2 \pi r^2}{L}$

The induced current is the result of the magnetic flux that couples (i.e. flows through) the wire loops in the solenoid and the outer sensing coil. This is Faraday's Law. The solenoid flux in one direction is essentially entirely contained within the solenoid.
View attachment 360418


So if the sensing coil wraps around the solenoid (sort of tightly*) the angle doesn't matter, the flux through it will be the same.

An analogy is if you were counting cars on a freeway that pass through an imaginary plane, it wouldn't matter if that plane was tilted, as long as every car had to pass through it somewhere.

* You don't want that outer coil to capture much of the returning flux. A gigantic outer coil might not see any net flux, it could all (mostly) cancel.

I understand

 

[songoku]

Because the magnetic field inside a very long solenoid of $N$ turns and length $L$ is approximated as $B=\mu_0 n I$ where $n=\dfrac{N}{L}$ is the number of turns per unit length.

For electric field in question (b), the correct answer is $-\frac{\mu_o abn}{2}$ or $-\frac{\mu_o abnN}{2}$?

Thanks

 

[kuruman]

For electric field in question (b), the correct answer is $-\frac{\mu_o abn}{2}$ or $-\frac{\mu_o abnN}{2}$?

What is the expression for the magnetic field $B$ inside a solenoid?
What is the expression for the flux through the single loop in this problem?

 

[haruspex]

For electric field in question (b), the correct answer is $-\frac{\mu_o abn}{2}$ or $-\frac{\mu_o abnN}{2}$?

Thanks

Since n=N/L, you are effectively asking whether it is $-\frac{\mu_o abN}{2L}$ or $-\frac{\mu_o abN^2}{2L}$.
Perhaps you meant should it be $-\frac{\mu_o abn}{2}$ or $-\frac{\mu_o abN}{2}$? If so, to put it another way, would the result be different if the solenoid were longer, but still the same winding density?
Well, what do you think?

 

[songoku]

Since n=N/L, you are effectively asking whether it is $-\frac{\mu_o abN}{2L}$ or $-\frac{\mu_o abN^2}{2L}$.
Perhaps you meant should it be is $-\frac{\mu_o abn}{2}$ or $-\frac{\mu_o abN}{2}$? If so, to put it another way, would the result be different if the solenoid were longer, but still the same winding density?
Well, what do you think?

I am actually asking whether it is $-\frac{\mu_o abN}{2L}$ or $-\frac{\mu_o abN^2}{2L}$ but I think now I know why I am wrong.

What is the expression for the magnetic field $B$ inside a solenoid?
What is the expression for the flux through the single loop in this problem?

I was mistaken when taking which object to consider. I thought I need to consider the number of turns of solenoid. It turns out I need to consider the loop.


Thank you very much for all the help and explanation kuruman, DaveE, haruspex