To find work done by time-varying magnetic field

In summary, the problem involves a magnetic field varying at a constant rate in a cylindrical region with a triangular loop lying within it. The question asks for the work done to move a unit positive charge from one side of the loop to the other. Using Faraday's law and the area of the triangle, the correct answer is (c).
  • #1
hermy
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Homework Statement



Magnetic field in the cylindrical region with its axis passing through O varies at a constant rate x. A triangular imaginary loop ABC, with AB = BC, is lying in this region as shown in figure. The work done to move unit positive charge from A to B along side AB is:

(a) r2 x

(b) [tex]\pi[/tex] r2 x

(c) r2/2 x

(d) 3r2/4 x


Homework Equations



emf induced in a circular loop of radius a due to time varying magnetic field = [tex]\pi[/tex] a2 x

work done in moving a charge q = qV

The Attempt at a Solution



Honestly, I have no clue how to go about it. Electric field is non-conservative, so we can't simply find the potential difference. When I tried to integrate E.dl , I could not find the angle between electric field and displacement. I feel pretty sure that we need to integrate, but how? A hint would be sufficient.

thanks in advance.
 

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  • #2
You can use Faraday's law: integral of E dl (induced electromotive force) is equal to the time rate of change of the magnetic flux through the loop. You will only need to calculate area of the triangle, so no integration is necessary.
 
  • #3
i got it. so the answer should be (c). thanks eloy.
 

What is work done by a time-varying magnetic field?

Work done by a time-varying magnetic field is the amount of energy transferred to a charged particle as it moves through the field. This energy transfer is due to the changing magnetic field inducing an electric field and causing the particle to experience a force.

How is work done by a time-varying magnetic field calculated?

The work done by a time-varying magnetic field can be calculated using the equation W = q∫E•ds, where W is the work done, q is the charge of the particle, E is the electric field induced by the magnetic field, and ds is the displacement of the particle.

What is the relationship between work done and the strength of the time-varying magnetic field?

The work done by a time-varying magnetic field is directly proportional to the strength of the field. This means that as the strength of the magnetic field increases, the amount of work done on the charged particle also increases.

Can work be done by a constant magnetic field?

No, work cannot be done by a constant magnetic field. A constant magnetic field does not induce an electric field, therefore there is no force acting on the charged particle and no work is done.

What are some real-life applications of work done by time-varying magnetic fields?

Work done by time-varying magnetic fields is used in many technological devices, such as generators, electric motors, and transformers. It is also essential in the production of electricity through the process of electromagnetic induction.

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