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.

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:

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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.

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Final Answer:

$$ \varepsilon = 2 B l v \sin\left(\frac{\theta}{2}\right) $$