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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2 \pi i a B$

Vedclass · Answer

(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?

Similar Questions

An arrangement of three parallel wires placed perpendicular to the plane of the paper,each carrying the same current $I$ in the same direction,is shown in the figure. The magnitude of the force per unit length on the middle wire $B$ is given by:

Two parallel wires in free space are $10\, cm$ apart and each carries a current of $10\, A$ in the same direction. The force one wire exerts on the other per metre of length is

Three long straight wires are carrying current as shown in the diagram. The magnetic force per unit length on wire $Q$ is $..... \times 10^{-7} \ N/m$.

Two parallel wires $1 \text{ m}$ apart carry currents of $1 \text{ A}$ and $3 \text{ A}$ respectively in opposite directions. The force per unit length acting between these two wires is

$A$ straight current-carrying conductor is placed such that the current flows out of the plane of the paper. The conductor is placed between the poles of two magnets,as shown in the figure. The conductor will experience a force in the direction towards:

Vedclass Test Series

Exam Paper Generator

Online Exam Module