In Fig. 𝐏 29.49 the loop is being pulled to the right at...

In Fig. $\mathbf{P} 29.49$ the loop is being pulled to the right at constant speed $v .$ A constant current $I$ flows in the long wire, in the direction shown. (a) Calculate the magnitude of the net emf $\mathcal{E}$ induced in the loop. Do this two ways: (i) by using Faraday's law of induction (Hint: See Exercise 29.7 ) an (ii) by looking at the emf induced in each segment of the loop due to its motion. (b) Find the direction (clockwise or counterclockwise) of the current induced in the loop. Do this two ways: (i) using Lenz's law and (ii) using the magnetic force on charges in the loop. (c) Check your answer for the emf in part (a) in the following special cases to see whether it is physically reasonable: (i) The loop is stationary; (ii) the loop is very thin, so $a \rightarrow 0 ;$ (iii) the loop gets very far from the wire.

Key Concepts

Evaluation of Special Cases for Physical Reasonableness

Checking limiting cases of a derived emf expression, such as when a system is stationary or when certain dimensions shrink to zero, is a method used to validate physical formulas. This approach ensures that the derived expressions behave as expected under extreme but physically relevant conditions, reinforcing the reliability of the theoretical model.

Faraday's Law of Induction

This principle states that an electromotive force (emf) is induced in a closed loop when there is a change in the magnetic flux through the loop. The magnitude of the induced emf is proportional to the rate at which the flux changes with time. It is a fundamental concept in electromagnetism that ties together time-varying magnetic fields and induced electrical currents.

Motional EMF

Motional emf arises when a conductor moves through a magnetic field, causing free charges in the conductor to experience a Lorentz force. This force separates charges and creates a potential difference along the conductor. It highlights that an emf can be induced not only by a changing magnetic field but also by the motion of conductors relative to a steady magnetic field.

Lenz's Law

Lenz's law provides the direction of the induced current in a loop, stating that the induced current will flow in such a way as to oppose the change in magnetic flux that produced it. This law is critical for determining the sign of the induced emf and ensuring the conservation of energy within electromagnetic systems.

Magnetic Force on Moving Charges

When charges move through a magnetic field, they experience a magnetic force described by the Lorentz force law. This concept explains how moving charges in a conductor can be deflected by a magnetic field, leading to the separation of charges and the generation of an emf. Analyzing the force on individual segments of a conductor helps in understanding the overall induced emf in moving loop configurations.

Transcript

00:01 We have an infinite wire which carries current capital i so the diagram for the infinite wire can be shown like this okay this is the diagram for the condition okay so this is our diagram and the this is our infinite wire okay which carries current capital i and this is our loop okay which has side a and b which is moving towards right with speed v okay so first of all let us suppose an element which has width of d x at the position of x from the wire okay this side is r from the wire to this end okay so now for the part a of the question uh we can determine the flux linked with the loop of vertical side b okay so at a distance x from the wire okay so we can write that flux t5 will be equals to mu not i divided by 2 pi x multiplied by area area will be of this width okay so this width has area of b multiplied by d x okay so the total flux linked with the loop of vertical side b at a distance small r from the wire okay and it can be written as phi equals to integration of d5 and that will be equals to integration of mu not i divided by 2 pi x b multiplied by d x from small r to r plus a this is the limit of the loop minimum limit is r and maximum limit will be r plus a because this side is a okay so now taking out the constant values so we can write mu not i b divided by 2 pi okay and integration from r to r plus a we get one by x multiplied by d x okay so from here after solving we get five flux flux i equals to mu not i b divided by two pi l n r plus a divided by r okay so this is the value of flux so now according to the ferradi's law of electroph electromagnetic induction we can write the emf and that will be equals to emf equals to minus d phi by dt.

02:33 Okay.

02:34 So now we can write that emfe will be equals to minus d by dt of phi that is equals to mu not divided by 2 pi, i b, ln, r plus a divided by r.

02:49 Okay.

02:50 So this is the equation.

02:51 Now from here after solving we get e equals to minus m.

02:55 Not by 2 pi i b this is the value constant values okay so we have taken out these values so now we can write that differentiation of this value will be r divided by r plus a multiplied by minus a by r square and d r d t okay this is the method of sequencing of differentiation okay so now dr by dt will be equals to velocity v so it can it will be equals to mu not ib divided by 2 pi and mu not i b and a divided by r r plus a multiplied by v okay so after rearranging these terms we can write mu not divided by 2 pi i a b divided by r r plus a multiplied by by v okay so this is the emf of the emf induced in the loop okay so now this is the value of answer for the part a of part one okay so now see to the part two which is second part of the problem so since emf is induced in horizontal sides are zero emf in the horizontal sides will be 0, horizontal sides equals to 0 and these sides does not cut a cut magnetic line of force.

04:35 So the emf in the vertical side which is near to the wire which is even, which can be written as mu not divided by 2 pi i by r multiplied by v multiplied by b.

04:50 And similarly, emf induced in the farther side of vertical side, this will be equals to e2 and this will be equals to mu not divided by 2 pi, i divide by this is capital i, i divide by r plus a because the distance will be r plus a multiplied by vb.

05:10 So the emf net in the loop will be e1 minus e2 and that will be equals to mu not divided by 2 pi.

05:20 I v b bracket 1 by r minus 1 by r plus a okay so from here after solving b get e equals to mu not divided by 2 pi i a b divided by r r plus a made flat by b okay which is the same result as obtained in the part one of the part a of the question okay so now moving to the part b of the problem okay and first be and first part in which we will determine the direction of current using lens law and from the lens law as the loop is moving away moving away then the flux decreases flux through the loop will decrease okay so to in the loop will oppose this change so it will induced it will induce a clockwise current clockwise current in the loop so that it can oppose the change.

06:22 So this is the answer for the question...

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