An electric current $I$ enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged 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

Question

Q. An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged 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 equal to
(a)

q v \frac{μ_{0} i}{2 a}

(b)

q v \frac{μ_{0} i}{2 πa}

(c)

q v \frac{μ_{0} i}{a}

(d) zero

Answer 1

Since the direction of forces are opposite due to the direction of current in loop i.e.

\frac{l}{2}

and

\frac{l}{2}

.

Answer 2

So force acting are same in magnitude but opposite in direction and so net force acting on the charge is zero.

The correct option is D ZeroSince the direction of forces are opposite due to the direction of current in loop i.e. l2 and l2.So force acting are same in magnitude but opposite in direction and so net force acting on the charge is zero.

The correct option is D Zero
Since the direction of forces are opposite due to the direction of current in loop i.e. $\frac{l}{2}$ and $\frac{l}{2}$ .
So force acting are same in magnitude but opposite in direction and so net force acting on the charge is zero.