Question

Three long wires, each carrying current $i$ are placed parallel to each other. The distance between $I$ and $II$ is $3 d$, between $II$ and $III$ is $4\, d$ and between $III$ and $I$ is $5 \,d$. Magnetic field at side of wire II is

• A. $\frac{5 \mu_{0} i}{24 \pi d}$
• B. $\frac{10 \mu_{0} i}{24 \pi d}$
• C. $\frac{15 \mu_{0} i}{24 \pi d}$
• D. $\frac{20 \mu_{0} i}{24 \pi d}$

Answer 1

Answer: (A)

At the wire II,
Magnetic field due to wire $I , B_{1}=\frac{\mu_{0}{ }^{j}}{2 \pi \cdot 3 d}$
Magnetic field due to wire III, $B_{2}=\frac{\mu_{0} j}{2 \pi \cdot 4 d}$
$B_{1}$ and $B_{2}$ are perpendicular.
Resultant magnetic field
$B_{R} =\sqrt{B_{1}^{2}+B_{2}^{2}}=\frac{\mu_{0} j}{2 \pi d} \sqrt{\frac{1}{3^{2}}+\frac{1}{4^{2}}} $
$=\frac{\mu_{0} i}{2 \pi d} \sqrt{\frac{4^{2}+3^{2}}{3^{2} \cdot 4^{2}}}$
$=\frac{5 \mu_{0} i}{2 \pi d \times 12}=\frac{5 \mu_{0} j}{24 \pi d} $