Question

A rectangular wire loop located in the plane of the page has a width L, and its length. X


A rectangular wire loop located in the plane of the page has a width L, and its length. X. is determined by the position a movable rail that forms the fourth side of the rectangle, as shown. The total electrical resistance of the wire loop is R, and an externally applied magnetic field, B, is directed out of the page. The rail is moving towards the right with speed v. Assume that the x direction is towards the right of the page, the y direction is towards the top of the page, and the z direction is out of the page. 

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Part (a) Using only the variables provided what is the magnitude of the magnetic flux at the instant when the rectangle has length x? - 

Part (b) Using only the variables provided, what is the magnitude of the induced EMF in the loop? 

Part (c) What is the direction of the EMF generated in the loop? 

Part (d) What is the magnitude of the induced current in the loop?  

Part(e) What external force, Fapplied, must be applied to the rail to keep it moving to the right at constant speed?

Part (f) Based upon the calculation of the force, Fapplied, calculated in the previous step, how much power must be put into the system in order to maintain the constant speed of the rail? 

Part (g) Considering the current calculated in a previous step, calculate the power dissipated through the resistor 

Part (h) Which statement best describes the flow of energy through the system?

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Answer #1

a)We know,

Flux = B×A

here, B = magnetic field.

A = area of the surface = (L×x)

So the flux in the rectangle= B×x×L

b) We know,

e = B.L.v.sin

here, e = induced emf.

v = velocity of the rod.

= angle between the velocity and magnetic field = 90°

So, the induced emf = B×L×v

c) According to Lenz law the induced emf always opposes the cause of the induction. So the resultant emf will be formed in clockwise direction. Then, the induced magnetic field will oppose the magnetic field of outside.

d) Induced current in the loop = (e/R)

=> The induced current in the loop = (B×L×v/R)

[As per guidelines, I'm bound to answer first 4 subparts of a question when more than 4 is posted together. Please post the other questions separately]

Please comment if you have any doubt and like if it helps.

Happy learning.

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