The shaded region in the diagram represents a uniform magnetic field that is dir

Question

The shaded region in the diagram represents a uniform magnetic field that is directed perpendicular to the page. (middot = +z and x = - z.) This field varies periodically in time: B_z(t) = B_max Sin(omega t), where B_max = 0.789 T, and t is measured in seconds. The frequency of this field's cycle is 0.564 Hz. There is a loop of conductive wire (shaped as a right triangle), with dimensions as shown, that has a total resistance of 0.213 Ohm. It is being moved into the field from left to right at a constant velocity of 1.25 m/s, without any twisting or bending. At the moment depicted here (t = 4.60 s), half of the loop's base is already in the field. Calculate the current in the loop (including its direction, cw or ccw) at the moment depicted here.

Explanation / Answer

Solution:

From the given details we need to calculate B.

BZ= (.789)*sin(2*3.14*.564*4.6)=.22T (since =2f)

The idea is to calculate total induced emf (V) and divide it by Resistance to get the current.

By Faraday’s law e.m.f. =B l v

We have calculated B already.

Now we need to calculate l.

Wkt, In a right angled triangle the ratio of adj. to opp. are constant.

Thus, 115/6.9=57.5/l i.e. l= 3.45m (since half of the base is 57.5m)

Velocity is given as 1.25m/s

And in the diagram given, I am not able to see the X or . indicating the directions, you will need to calculate the values for clock wise and anti-clockwise accordingly.

Thus,e.m.f=.94875 V

And I= e.m.f./Resistance= .94875/.213= 4.454A

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