Questions

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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detailed solution

Correct option is C

Using, impulse = change in linear momentum, we have∫Fdt=mv or ∫(iBl)dt=mvor Blq = m v as ∫idt=q∴ q=mvBl

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