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

Q: 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

d (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 $O$ ${B_{net}} = \frac{{{\mu _0}}}{{4\pi }} \times \frac{2}{x} \times ({i_1} - {i_2})$ $ \Rightarrow 10 \times {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 \times {10^{ - 6}} = \frac{{{\mu _0}}}{{4\pi }}.\frac{2}{x}({i_1} + {i_2})$ ..... $(ii)$ Dividing equation $(ii)$ by $(i)$ $\frac{{{i_1} + {i_2}}}{{{i_1} - {i_2}}} = \frac{4}{1} \Rightarrow \frac{{{i_1}}}{{{i_2}}} = \frac{5}{3}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$3 : 4$
B$11 : 7$
C$7 : 11$
D$5 : 3$

Diffcult

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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