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An electric current $\imath$ enters and leaves a uniform circular wire of radius $ \Re$ through diametrically opposite points. A charged particle q moving along the axis of the circular wire passes through the centre at speed $v$. The magnetic force acting on the particle when it passes through the centre is

$\begin {array} {1 1} (a)\;QV\large\frac{\mu_0i}{2 \pi a} & \quad (b)\;QV\large\frac{\mu_0i}{4 a} \\ (c)\;QV\large\frac{\mu_0i}{2 a} & \quad (d)\;Zero \end {array}$


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 As the current is entering and exiting from two diametrically opposite points of a circular coil, the currents in the two semicircular sections are in the opposite direction.
The fields due to the two semi-circular sections at the centre is in the opposite direction.
hence net feild is zero
Ans : (d)


answered Feb 22, 2014 by thanvigandhi_1
edited Sep 18, 2014 by thagee.vedartham

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