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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 -

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Question

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 ____________.

Options

  • zero

  • Bvπr2/2 and P is at higher potentia

  • πrBv and R is at higher potential

  • 2rBv and R is at higher potential

MCQ
Diagram

Solution

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 2rBv and R is at higher potential.

Explanation:

`"e" = "B"l_"eff" "v" ("where"  l_"eff" ="Diameter")`

`= "B" (2"r")"v" = 2"rBv"`

and R is at higher potential by Fleming's right hand rule.

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Lenz's Law
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