Question

A square loop PQRS carrying a current of 6.0 A is placed near a long wire carrying 10 A as shown in figure. (a) Show that the magnetic force acting on the part PQ is equal and opposite to the part RS. (b) Find the magnetic force on the square loop.
Figure

Answer 1

Explanation:

(a) To show: the magnetic force acting on the part PQ is equal and opposite to the part RS.

Step 1:

Given data in the question:

Loop Current $i_1$ = 6 A

Wire Current $i_2$ = 10 A

Now consider an element at a distance x from the wire on PQ of width dx.  

Step 2:

Force on the entity is given by

$d F=\frac{\mu_{o} i_{1} i_{2}}{2 \pi x} d x$

Step 3:

Force acting on part PQ is indicated by

$F_{P Q}=\frac{\mu_{o} t_{1} t_{2}}{2 \pi x} \int_{1}^{3} \frac{d x}{x}$

$=2 \times 10^{-7} \times 6 \times 10[\ln x]_{1}^{3}$

$=120 \times 10^{-7} \ln 3 N$

Likewise

$F_{R S}=\frac{\mu_{o} i_{1} i_{2}}{2 \pi x} \int_{3}^{1} \frac{d x}{x}$

$=120 \times 10^{-7} \times \ln \frac{1}{3}$

$=-120 \times 10^{-7} \ln 3 N$

All forces are of equal magnitude but in the opposite direction.  Hence proved

(b) To find: The magnetic force on the square loop

Step 1:

The intensity of the magnetic field due to SP wire is given by

$B=\frac{\mu_{0} i_{2}}{2 \pi r}$

Step 2:

Force is given on part by SP

$F_{S P}=i_{1} B l$

$=\frac{\mu_{o} t_{1} t_{2}}{2 \pi} \frac{l}{r}$

$=\frac{\mu_{o} i_{1} i_{2}}{2 \pi} \frac{2}{1}$

(Turn right)

Step 3:

Force on the part of the RQ

$F_{R Q}=i_{1} B l$    

$=\frac{\mu_{o} i_{1} i_{2}}{2 \pi} \frac{l}{r}$

$=\frac{\mu_{o} t_{1} t_{2}}{2 \pi} \frac{2}{3}$

(left  Towards)

Step 4:

The net force on the loop is thus given by the

$F_{n e t}=F_{S P}-F_{R Q}$      

$=\frac{\mu_{o} i_{1} i_{2}}{2 \pi}\left(\frac{2}{1}-\frac{2}{3}\right)$  

$=2 \times 10^{-7} \times 6 \times 10 \times \frac{4}{3}$

$=16 \times 10^{-6} N$

( right Towards)

Therefore the magnetic force on the square loop is $16 \times 10^{-6} N$

Answer 2

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