Question

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 AOB​

Answer 1

Explanation:

Some scales, such as the traditional beam balance, measure mass directly by counteracting the downward force by another equal downward force on the opposite side of the pivot. On a beam balance, you add enough mass to the opposite pan to counterbalance the unknown mass. No force is measured.

Answer 2

Given: Length is equal to x

Find: We have to calculate the EMF induced between the two free ends

Step by Step Solution:

Consider the circuit as being closed externally,  

AD = x

So, the area of the rectangular part ABCD is

$A = 2xl sin(\dfrac{\theta}{2})$

As the angle moves towards right, the flux of the magnetic field through this rectangular area decreases at the rate

$\dfrac{d\phi}{dt} = B\dfrac{dA}{dt} = 2Blvsin(\dfrac{\theta}{2})$

This is also the rate of decrease of the flux through the closed circuit.

so the induced EMF is,

$\epsilon = 2Blvsin(\dfrac{\theta}{2})$

As, the EMF is induced solely because of the movement of the angle, this is the EMF induced between its ends.