Question

a conducting rod OA of length 2l is rotated about its end O with a angular velocity omega in a uniform magnetic field B directed perpendicular to the plane of rotation. the potential difference across its two ends is
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Answer 1

i could not understand

Answer 2

Given:

Given that, a conducting rod OA whose length is 2l is rotated about its end O with a angular velocity in a uniform magnetic field B which is directed perpendicular to the plane of rotation.

To find:

We need to find the potential difference across the two ends of the rod.

Solution:

Let the rod of length 2l is moved with angle $\triangle \theta$, then

We know that, Angle =  arc / radius

$\triangle \theta = \frac{\text{arc}}{2l}$ [As radius is the length of the rod]

$\text{arc} = 2l \times \triangle \theta$

Now, area swept in time t is

$\[\Delta A=\frac{1}{2}\times 2l\times \text{arc}\]$

       $\[=\frac{1}{2}\times 4{{l}^{2}}\times \Delta \theta$

Now, we know

$\Delta \phi =B\Delta A$

     $\[=\frac{1}{2}\times B\times 4{{l}^{2}}\times \Delta \theta \]$  

Now,

E.m.f $=\frac{\Delta \phi }{\Delta t}$

        $=\[\frac{1}{2}\times B\times 4{{l}^{2}}(\frac{\Delta \phi }{\Delta t})$

         $=\frac{1}{2}\times B\times 4{{l}^{2}} \omega$    

Therefore, E.m.f $\[=2B{{l}^{2}}\omega$.