Question

In the figure, a long straight wire carries a current $$i_1=30\hspace{0.05cm}A$$
and a rectangular loop carries current $$i_{2}=20\hspace{0.05cm}A$$. Take the dimensions to be $$a=1\hspace{0.05cm}cm,b=8\hspace{0.05cm}cm$$ and $$L=30\hspace{0.05cm}cm$$. In unit-vector notation, what is the net force on the loop due to $$i_{1}$$?

Answer 1

$$Given$$ :- $$i_1 = 30A$$ , $$i_2 = 20A$$

$$a= 1cm$$ , $$b= 8cm$$ and $$L= 30cm$$

$$To \space Find$$ :- Net force on loop due to $$i_1$$

$$ Solution$$ :- We know, field $$(B)$$ due to current $$I$$ is given by ,

$$B = \dfrac{\mu_0I}{2πr}$$

Now , Let's divide the loop into 4 parts i.e,

$$1-2\space ,\space 2-3\space ,\space 3-4\space ,\space and \space 4-1$$

$$\therefore$$ Field at $$1-2$$ $$(\vec B_{1-2}) = \dfrac{\mu_0i_1}{2πa}$$ ( The direction is into plane of paper )

Similarly at $$3-4$$ $$(\vec B_{3-4}) = \dfrac{\mu_0i_1}{2π(a+b)}$$ ( The direction is into the plane )

Now , The force $$\vec F$$ on a wire of length $$L$$ carrying current $$I$$ in

magnetic field of strength $$\vec B$$ (constant) is given by ,

$$\vec F = L(\vec I × \vec B)$$

$$\vec F_{1-2} = L( i_2 × B_{1-2})$$ ( The direction is up i.e , along $$\hat{j}$$ )

$$=(\dfrac{\mu_0i_1i_2L}{2πa})\space \hat j$$

Similarly $$\vec F_{3-4} = (- \dfrac{\mu_0i_1i_2L}{2π(a+b)})\space \hat j$$ ( The direction is down i.e , along $$-\hat j$$ )

And force on part $$2-3 $$ and $$4-1 $$ is equal but opposite , hence cancelled

$$\implies$$ $$\vec F_{2-3} + \vec F_{4-1} = 0$$

$$\therefore $$ $$F_{net}= F_{1-2}+F_{3-4}$$

$$=(\dfrac{\mu_0i_1i_2L}{2πa} - \dfrac{\mu_0i_1i_2L}{2π(a+b)})$$ $$\hat j$$

$$=[\dfrac{\mu_0i_1i_2L}{2π}(\dfrac{1}{a}-\dfrac{1}{a+b})]\space \hat j$$

$$= [\dfrac{4π×10^{-7}×30×20×30×10^{-2}}{2π}(\dfrac{800}{9})] \hat j$$

$$=\underline{ (3.2×10^{-3} \space N)\hat j}$$

Hence , $$\underline{Net \space force \space is \space 3.2× 10^{-3} \space N \space towards \space the \space straight \space wire.}$$