A circular loop of radius a , carrying a current I ,is placed in a 2D magnetic field. The centre of the loop coincides with the centre of the field. The strength of magnetic field at the periphery of loop is B. Find the magnetic force on the wire?
This solution does not appear to be correct as the magnetic field assumed is not 2D but a 3D.
It may be ignored.
We know that Magnetic Force on a Current-Carrying Wire is perpendicular to both the wire and the magnetic field. The direction of magnetic force is given by the right hand rule or by cross product of the two vectors. It comes from the expression for
As total Magnetic Force is the Sum of Forces on all charges in the wire, it can be shown for a current
For a circular loop of radius
Lets consider infinitesimal length
It is given that the centre of the loop coincides with the centre of the 2D magnetic field. The strength of magnetic field at the periphery of loop is
We can therefore assume that the loop of wire is located in a uniform magnetic field
To find the total magnetic force we need to find the integral of equation (2) over the loop. We get
We see that infinitesimal vector lengths
Equation (3) becomes
For each infinitesimal length of the current carrying loop, we can locate a diametrically opposite infinitesimal length carrying current in the opposite direction.
A circular loop of radius a, carrying a current I,is placed in a 2D magnetic field
Let us consider an infinitesimal length
As the angle between
By Fleming's Left Hand Rule or by Rule of cross product of two vector we can easily determine the direction of force experienced by the considered small length segment (ds) of the loop. This force will be directed perpendicular to the plane of the loop as shown in the figure.
If we are to find the total magnetic force on the loop , we are to add up the all magnetic forces acting on the all infinitesimal length segments constituting the loop.
It can be done by integrating equation (1) all over the circumference(