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An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points.Acharged particle q moving along the axis of the circular wire passes through its centre at speed v.the magnetic force acting on the particle when it passes through the centre has a magnitude -------. An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points.Acharged particle q moving along the axis of the circular wire passes through its centre at speed v.the magnetic force acting on the particle when it passes through the centre has a magnitude -------.
An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points.Acharged particle q moving along the axis of the circular wire passes through its centre at speed v.the magnetic force acting on the particle when it passes through the centre has a magnitude -------.
qv× μ0⁄2a
When the current enters in the circular loop, it breaks into 2 parts one in each semicircular arm. By using the right hand pal rule, we get to know that the direction of magnetic field is equal and opposite at the centre and since we use the formula f=q(v×B) to find the resultant at the centre, the opposite direction would cancel the two forces. Therefore there is no net effect of the force.
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