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Q.

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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a

2mvBl

b

Bl2mv

c

mvBl

d

Blv2m

answer is C.

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Detailed Solution

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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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