Question

Two long parallel current carrying conductors $30cm$ apart having current each of $10A$ in opposite direction, then value of magnetic field at a point of $10cm$ between them, from any of the wire.

• A. $1\times {10}^{-5}T$
• B. $6\times {10}^{-5}T$
• C. $1.5\times {10}^{-5}T$
• D. $3\times {10}^{-5}T$

Answer 1

The correct option is A $1\times {10}^{-5}T$

Given that,

Distance $d=30cm$

Current $I=10A$

Suppose that $A$ and $B$ is Two long parallel current carrying conductors

We know that,

The magnetic field at a point of 10 cm

$B_A=\frac{{\mu }_0\times I}{2\pi d}$

$B_A=\frac{4\pi \times {10}^{-7}\times 10}{2\pi \times 0.1}$

$B_A=2\times {10}^{-5}T$

Now, the magnetic field at a point of 20 cm

$B_B=\frac{{\mu }_0\times I}{2\pi d}$

$B_B=\frac{4\pi \times {10}^{-7}\times 10}{2\pi \times 0.2}$

$B_B=1\times {10}^{-5}T$

Now the net magnetic field is

$B=B_A-B_B$

$B=2\times {10}^{-5}-1\times {10}^{-5}$

$B=1\times {10}^{-5}T$

Hence, the magnetic field is $1\times {10}^{-5}T$