0.12 Magnetic force on a conductor  (Page 4/5)

Problem : An irregular shaped flexible wire loop of length “L” is placed in a perpendicular and uniform magnetic field “B” as shown in the figure below (The magnetic force represented by filled circle is perpendicular and out of the plane of drawing). Determine the tension in the loop if a current “I” is passed through it in anticlockwise direction.

An irregular shaped flexible wire loop in magnetic field

Solution : The wire loop is flexible. There would be tension, provided the loop elements experience magnetic force in outward direction at all points on it. Applying Right hand thumb rule for any small segment of the loop, we find that the wire is indeed subjected to outward magnetic force. Clearly, the loop expands to become a circular loop. The radius of the circle is given by :

An irregular shaped flexible wire loop in magnetic field

$$2\pi r=L$$ $$\Rightarrow r=\frac{L}{2\pi }$$

In order to determine tension in the wire, we consider a very small element of the circular loop. Let the loop element subtends an angle dθ at the center. Let “T” be the tension in the wire. It is clear that components of tension in the downward direction should be equal to magnetic force on the small wire element.

The tension in the circular loop carrying current

$$2T\sin \frac{đ\theta }{2}=F_M$$

Since loop element is very small, we approximate as :

$$\sin \frac{đ\theta }{2}\approx \frac{đ\theta }{2}$$

Further, we can consider the small loop element to be a straight wire for the calculation of magnetic force. Now, the magnetic force on the loop element is :

$$F_M=IBđL=IBrđ\theta$$

Substituting in the equilibrium equation,

$$\Rightarrow 2T\frac{đ\theta }{2}=IBrd\theta$$ $$\Rightarrow T=IBr$$

Again substituting for the radius of circle, we have :

$$\Rightarrow T=\frac{ILB}{2\pi }$$