A 25.0 cm times 100.0 cm rectangular loop of wire situated so that the side to t

Question

A 25.0 cm times 100.0 cm rectangular loop of wire situated so that the side to the right of the broken line is in a uniform upward(out of the page) magnetic field of intensity 0.405 T and the left side (of the dotted line) is not. the total resistance of the loop is 0.315 Ohm. Calculate the force required to pull the loop from the field (to the left) at a constant velocity of 25.0m/s. Neglect gravity. In your analysis, determine (a) the direction if the induced current in the loop and (b) give the direction of the forces on the four side of the loop. State why the forces, on the longer sides of the loop, do not influence the constant motion of the loop to the left.

Explanation / Answer

applied magnetic field (B) = 0.405T

resistance of the lop   (R)= 0.315 ohm

loop moving velocity (V) = 25m/s

length of the loop (l)= 25cm=0.25m     (the magnetic field effect in shorter length only)

the force require to full the loop through this constant velocity is given by,

                                               Fpull=Fmag=(lVB)^2/R

                                                                =(0.25*25*0.405)^2/0.315 N

                                                                 =8.92 N

(a) the direction of the induced current is only in the shorter sides of the loop only, since the applied magnetic field doesnot effect the longer sides of the loop as these are in parrlel direction to applied magnetic field.thus ,these two forces will cancel to each other.finally the conrtibution of forces contributon due to from shorte sides of the loop only.  

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