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(A) The magnetic field $B$ produced by the isolated north pole at any point on the circular loop is directed radially outward from the centre.The curr

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Nearly $2 \pi a i B$ perpendicular to the plane of the wire

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(A) The magnetic field $B$ produced by the isolated north pole at any point on the circular loop is directed radially outward from the centre.The current $i$ flows along the circumference of the loop. For any small element $dl$ of the wire,the magnetic force $dF$ is given by $dF = i(dl 	imes B)$.Since the magnetic field $B$ is radial and the current element $dl$ is tangential to the circle,the angle between $dl$ and $B$ is $90^\circ$ at every point.Thus,the magnitude of the force on a small element is $dF = i dl B \sin(90^\circ) = i B dl$.By Fleming's left-hand rule,the direction of this force is perpendicular to the plane of the loop (directed into or out of the plane depending on the direction of current).Since the force on every element is directed in the same direction (perpendicular to the plane),the total force $F$ is the integral of $dF$ over the entire circumference:$F = \int dF = \int i B dl = i B \int dl = i B (2 \pi a) = 2 \pi a i B$.Therefore,the total force is $2 \pi a i B$ perpendicular to the plane of the wire.

Suppose an isolated north pole is kept at the centre of a circular loop carrying an electric current $i$. The magnetic field due to the north pole at a point on the periphery of the wire is $B$. The radius of the loop is $a$. The force on the wire is

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