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A conducting rod of mass m and length l is placed over a smooth horizontal surface. A uniform magnetic field B is acting perpendicular to the rod. Charge q is suddenly passed through the rod and it acquires an initial velocity v on the surface, then q is equal to

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$\frac{2 mv}{Bl}$

$\frac{mv}{Bl}$

$\frac{Bl}{2 mv}$

$\frac{Blv}{2 m}$

answer is C.

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Detailed Solution

using, impulse = change in linear momentum, we have

$\int Fdt = mv or \int (iBl) dt = mv$

$or Blq = mv (as \int idt = q)$

$∴ q = \frac{mv}{Bl}$

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