Q.

A structure formed from two straight conducting rods forming a right angle, is placed in a uniform magnetic field with B = 0.40 T is directed out of the page. A straight conducting bar in contact with the structure, starts at the vertex at time t = 0 and moves with a constant velocity of 5.0 m/s along them. The emf (in V) around the triangle atthat time t=4.0 s is __________.

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answer is 80.

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Detailed Solution

The perpendicular length of the triangular area enclosed by the structure and bar is the same as the distance travelled by it in time td=vt, where v=5.0 m/s.Here the "base" of that triangle (the distance between the intersection points of the bar with the structure) is 2d. Thus, the area of the triangle isA=12( base )( height )=12(2vt)(vt)=v2t2Since the field is a uniform B = 0.350 T, then the magnitude of the flux (in SI units) isϕB=BA=(0.40)(5.0)2t2=10.0t2The magnitude of the emf is the (absolute value of) Faraday's law:ε=dϕBdt=10.0dt2dt=20.0t At t=4.0s, this yields ε=80.0V
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A structure formed from two straight conducting rods forming a right angle, is placed in a uniform magnetic field with B = 0.40 T is directed out of the page. A straight conducting bar in contact with the structure, starts at the vertex at time t = 0 and moves with a constant velocity of 5.0 m/s along them. The emf (in V) around the triangle atthat time t=4.0 s is __________.