Induced Electric Field In Circuit With Varying Resistance

[PumpkinCougar95]

Homework Statement

This is irodov's problem #310 in electrodynamics

A long solenoid of cross-sectional radius a has a thin insulated wire ring tightly put on its winding; one half of the ring has the resistance $\beta$ times that of the other half. The magnetic induction produced by the solenoid varies with time as B = bt, where b is a constant. Find the magnitude of the electric field strength in the ring.

Homework Equations

Provided solution:
https://drive.google.com/file/d/0B2jHGkWhC0E2eWtIOXV2SWFVdmM/view?usp=sharing

The Attempt at a Solution

Why have we not used the formula :

$E = \frac {r}{2} \frac {dB} {dt}$ for the Electric field , where r is the radius of the circle.
Why is the electric field dependent on the resistance?

Also, How can charges just accumulate in a wire? Why doesn't this happen in other cases?

Lastly, The solution states that the current in the 2 sections must be different, yet it has taken the current to be the same in its first two equations, why is this?

Thanks for your help!

 

[cnh1995]

Also, How can charges just accumulate in a wire?

They accumulate on the surface of the wire to aid the electric field in the part of higher resistance and oppose the electric field in the part of lower resistance, so as to make the currents equal in both the parts. It is a series circuit after all.

Lastly, The solution states that the current in the 2 sections must be different,

The currents are different initially. This causes the charges to accumulate at the junction of the two resistances and the currents become equal in very short time (typically of the order of 10^-12 s or smaller).

 

[PumpkinCougar95]

Thanks for Replying
But i thought that whenever there is a changing B field it creates a E field by the equation $E = \frac {r}{2} \frac {db}{dt}$ irrespective of the circuit in question. Then how is it that in this case, the E field is dependent upon the resistance? Is it that the E field the question is asking about an electrostatic field created by those induced charges?
Also, how do you know the charges will be on the surface?

 

[cnh1995]

Also, how do you know the charges will be on the surface?

Initially, the currents (or flows of charges) will be different in both the parts. This means there is a momentary charge imbalance i.e. in one part, there is an excess of electrons while in the other part, there is a deficiency of electrons. Where does the extra charge in a conductor go? It has to accumulate on the surface. This accumulation stops only when both the currents are same, i.e. no charge imbalance.

But i thought that whenever there is a changing B field it creates a E field by the equation E=r2dbdtE=r2dbdtE = \frac {r}{2} \frac {db}{dt} irrespective of the circuit in question.

First, the formula doesn't look correct. Area of the loop is pi*r².

Yes, the "induced" electric field is independent of the resistance and by symmetry, it should be same in both the halves of the circuit. But this gives rise to charge imbalance and in response, the surface charges rearrange themselves until there is no charge imbalance. This means, while the "induced" emfs are independent of the resistance, the "net" electric fields in both the parts strongly depend on the resistances of the parts.

This happens in circuits without varying B-fields as well. Look up 'surface charge feedback mechanism' in circuits. It will give you some insight about the concept of voltage drop and Kirchhoff's rules. It is very interesting.