A ring of radius a carries a current i. A mag | vedclass

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(D) The magnetic force on a small current element $dl$ is given by $dF = i(dl 	imes B)$.In this case,the current $i$ flows along the circumference of

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$2 \pi i a B$

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(D) The magnetic force on a small current element $dl$ is given by $dF = i(dl 	imes B)$.In this case,the current $i$ flows along the circumference of the ring,and the magnetic field $B$ is directed radially outward.At any point on the ring,the current element $dl$ is tangential to the ring,while the magnetic field $B$ is radial.Therefore,the angle between $dl$ and $B$ is $90^\circ$.The magnitude of the force on a small element is $dF = i \cdot dl \cdot B \cdot \sin(90^\circ) = i B dl$.The direction of this force,by the right-hand rule,is perpendicular to both the tangent and the radial direction,which is perpendicular to the plane of the ring (out of the plane).Since the magnetic field is symmetric and directed radially outward everywhere,the forces on diametrically opposite elements will be equal in magnitude and point in the same direction (out of the plane).However,if we consider the vector sum of forces around the entire ring,for every element $dl$,the force $dF$ is directed perpendicular to the plane of the ring.Integrating $dF$ over the entire circumference: $F = \int dF = \int i B dl = i B \int dl = i B (2 \pi a)$.Wait,let's re-evaluate: The force $dF$ on an element $dl$ is $i(dl 	imes B)$. Since $dl$ is tangential and $B$ is radial,$dl 	imes B$ points perpendicular to the plane of the ring. Since this direction is the same for all elements,the forces add up.$F = \int i B dl = i B (2 \pi a) = 2 \pi i a B$.

$A$ ring of radius $a$ carries a current $i$. $A$ magnetic field $B$ is present at the center of the ring,directed radially outward in the plane of the ring. What is the net magnetic force on the ring?

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