Question

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

• 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)

$\frac{5{\mu }_{0}i}{24\pi d}$

According to right hand rule, magnetic field due to III will be upward and due to I, it will be horizontal leftward.

Now using the standard formula for magnetic field as:

$M=\frac{{\mu }_{0}i}{2\pi d}$

Magnetic field upward is:

$M=\frac{{\mu }_{0}i}{8\pi d}$................(1)

Magnetic field leftward is:

$M=\frac{{\mu }_{0}i}{6\pi d}$..................(2)

Since both the magnetic field are mutually perpendicular, so we will do the vector sum as under-root of sum of square of each term of equation 1 and 2, we get:

$M=\frac{5{\mu }_{0}i}{24\pi d}$