Question

Two long straight wires are parallel and carry current in the same direction.The current are 8.0 A and 12.0 A and the wires are separated by 0.4 cm .The magnetic field (in Tesla ) at a point midway between the wires is

• A. 0
• B. $4.0\times {10}^{-4}$
• C. $8.0\times {10}^{-4}$
• D. $20\times {10}^{-4}$

Answer 1

Answer: (B)

$4.0\times {10}^{-4}$

Magnetic field at O due to wire 1, $B_1=\frac{{\mu }_o{I}_1}{2\pi d}$ into the paper
where $d=0.2cm=0.002m$
$\therefore$ $B_1=\frac{{\mu }_o\left(8\right)}{2\pi \left(0.002\right)}$ into the paper
Magnetic field at O due to wire 2, $B_2=\frac{{\mu }_o{I}_2}{2\pi d}$ out of the paper
$\therefore$ $B_2=\frac{{\mu }_o\left(12\right)}{2\pi \left(0.002\right)}$ out of the paper
Thus net magnetic field at O $B_O=B_2-B_1=\frac{{\mu }_o(12-8)}{2\pi \left(0.002\right)}$ (out of paper)
$\therefore$ $B_O=\frac{(4\pi \times {10}^{-7})\left(4\right)}{2\pi \left(0.002\right)}=4.0\times {10}^{-3}T$