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  • Solve this problem A thin semicircular conducting ring (PQR) of radius 'r' is falling with its plane vertical in a horizontal magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is v, is:

A thin semicircular conducting ring (PQR) of radius 'r' is falling with its plane vertical in a horizontal magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is v, is:

  • Option 1)

    zero

  • Option 2)

    \frac{Bv\pi r^{2}}{2}  and  P is at higher potential

  • Option 3)

    \pi rBv and R is at higher potential

  • Option 4)

    2rBv and R is at higher potential

 

Answers (1)

best_answer

 

Motional EMF -

varepsilon = Blv
 

- wherein

B ightarrow magnetic field

l ightarrow length

u ightarrow velocity of u perpendicular to uniform magnetic field.

 

 Potential developed depends on displacement between end points i.e. 2R

\therefore \varepsilon mf=Bvl

\Rightarrow Bv.2r=2Bvr

point R will be at higher potential.


Option 1)

zero

This is incorrect option

Option 2)

\frac{Bv\pi r^{2}}{2}  and  P is at higher potential

This is incorrect option

Option 3)

\pi rBv and R is at higher potential

This is incorrect option

Option 4)

2rBv and R is at higher potential

This is correct option

Posted by

divya.saini

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