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 at
that time t=4.0 s is __________.

Question

Infinity Learn · Accepted Answer

The perpendicular length of the triangular area enclosed by the structure and bar is the same as the distance travelled by it in time t

d=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 is

$A = \frac{1}{2} (base) (height) = \frac{1}{2} (2 v t) (v t) = v^{2} t^{2}$

Since the field is a uniform B = 0.350 T, then the magnitude of the flux (in SI units) is

$ϕ_{B} = B A = (0.40) (5.0)^{2} t^{2} = 10.0 t^{2}$

The magnitude of the emf is the (absolute value of) Faraday's law:

$\begin{array}{l} ε = \frac{d ϕ_{B}}{d t} = 10.0 \frac{d t^{2}}{d t} = 20.0 t \\ At t = 4.0 s, this yields ε = 80.0 V \end{array}$

Magnetic flux and induced emf

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