Question:pq02
An angle AOB made of a conducting wire moves along its bisector through a magnetic field B as suggested by the figure. Find the emf induced between the two free ends if the magnetic field is perpendicular to the plane of the angle.
Figure 1:
Figure 2:
Text Solution:
Consider the circuit as being closed externally as shown in the second figure.
If AD = x, the area of the rectangular part ABCD is given by:
$$ A = 2x l \sin\left(\frac{\theta}{2}\right) $$
As the angle moves towards the right with velocity v, the flux of the magnetic field through this area decreases at the rate:
$$ \frac{d\Phi}{dt} = B \frac{dA}{dt} = 2 B l v \sin\left(\frac{\theta}{2}\right) $$
This is also the rate of decrease of the flux through the closed circuit.
So, by Faraday's law of electromagnetic induction, the induced emf is:
$$ \varepsilon = 2 B l v \sin\left(\frac{\theta}{2}\right) $$
As the emf is induced solely because of the motion of the angle, this is the emf induced between its ends.
Final Answer:
$$ \varepsilon = 2 B l v \sin\left(\frac{\theta}{2}\right) $$
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