Two parallel long wires carry currents i_1 an | vedclass

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Question

(d) Initially when wires carry currents in the same direction as shown.
Magnetic field at mid point $O$ due to wires $1$ and $2$ are respectively
${

Vedclass · Accepted Answer

$5 : 3$

Vedclass · Answer

(d) Initially when wires carry currents in the same direction as shown.
Magnetic field at mid point $O$ due to wires $1$ and $2$ are respectively
${B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_1}}}{x} \otimes $
and ${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_2}}}{x}\, \odot$ 
Hence net magnetic field at $O$ ${B_{net}} = \frac{{{\mu _0}}}{{4\pi }} 	imes \frac{2}{x} 	imes ({i_1} - {i_2})$
$ \Rightarrow 10 	imes {10^{ - 6}} = \frac{{{\mu _0}}}{{4\pi }}.\frac{2}{x}({i_1} - {i_2})$ ..... $(i)$
If the direction of $i_2$ is reversed then
${B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_1}}}{x} \otimes $
and ${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_2}}}{x} \otimes $
So ${B_{net}} = \frac{{{\mu _0}}}{{4\pi }}.\frac{2}{x}({i_1} + {i_2})$
$ \Rightarrow 40 	imes {10^{ - 6}} = \frac{{{\mu _0}}}{{4\pi }}.\frac{2}{x}({i_1} + {i_2})$ ..... $(ii)$
Dividing equation $(ii)$ by $(i)$ $\frac{{{i_1} + {i_2}}}{{{i_1} - {i_2}}} = \frac{4}{1} \Rightarrow \frac{{{i_1}}}{{{i_2}}} = \frac{5}{3}$

Two parallel long wires carry currents $i_1$ and $i_2$ with ${i_1} > {i_2}$. When the currents are in the same direction, the magnetic field midway between the wires is $10\, \mu T$. When the direction of $i_2$ is reversed, it becomes $40 \,\mu T$. the ratio ${i_1}/{i_2}$ is

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