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

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) (5 μ0 i/24 π d) (B) (10 μ0 i/24 π d) (C) (15 μ0 i/24 π d) (D) (20 μ0 i/24 π d)

Tardigrade Admin · Accepted Answer

At the wire II,
Magnetic field due to wire

I, B_{1} = \frac{μ _{0} ^{j}}{2 π \cdot 3 d}

Magnetic field due to wire III,

B_{2} = \frac{μ _{0} j}{2 π \cdot 4 d}

B_{1}

and

B_{2}

are perpendicular.
Resultant magnetic field

B_{R} = B_{1}^{2} + B_{2}^{2} = \frac{μ _{0} j}{2 π d} \frac{1}{3 ^{2}} + \frac{1}{4 ^{2}}

= \frac{μ _{0} i}{2 π d} \frac{4 ^{2} + 3 ^{2}}{3 ^{2} \cdot 4 ^{2}}

= \frac{5 μ _{0} i}{2 π d \times 12} = \frac{5 μ _{0} j}{24 π d}

Q. 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

$\frac{5 μ _{0} i}{24 π d}$

$\frac{10 μ _{0} i}{24 π d}$

$\frac{15 μ _{0} i}{24 π d}$

$\frac{20 μ _{0} i}{24 π d}$

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

24πd5μ0​i​

24πd10μ0​i​

24πd15μ0​i​

24πd20μ0​i​

Q. 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

$\frac{5 μ _{0} i}{24 π d}$

$\frac{10 μ _{0} i}{24 π d}$

$\frac{15 μ _{0} i}{24 π d}$

$\frac{20 μ _{0} i}{24 π d}$