Question

If magnetic field due to current I in figure (i) at the centre of cube is B, then magnetic field due to current I at the centre of same cube in figure (ii) is :

• A. $B_0$
• B. $3B_0$
• C. $B_0\sqrt{3}$
• D. $B_0\sqrt{2}$

Answer 1

Answer: (C)

$B_0\sqrt{3}$

The magnetic field at a perpendicular distance a from $O$ due to one arms is,

$=\frac{{\mu }_{0}I}{4\pi \sqrt{2}a}2\times \frac{1}{\sqrt{3}}(\hat{j}\times \frac{(-\hat{i}+\hat{k})}{\sqrt{2}})$

$=\frac{{\mu }_{0}I}{4\pi \sqrt{3}a}(\hat{k}+\hat{i})$

$\therefore$ Net magnetic field at the centre in figure (i)

$B_0=4\times \frac{{\mu }_{0}I}{4\pi \sqrt{3}a}=\frac{{\mu }_{0}I}{\sqrt{3}\pi a}$

The magnetic field at the centre due to two parallel wire is

$B_1=2\times \frac{{\mu }_{0}I}{4\pi \sqrt{2}a}\frac{\sqrt{2}}{\sqrt{3}}\times 2=\frac{{\mu }_{0}I}{\sqrt{3}\pi a}$

$\therefore$ Net field in figure (ii) is due to three pairs, all in perpendicular directions.

$\therefore B_{net}=\sqrt{3}\times \frac{{\mu }_{0}I}{\sqrt{3}\pi a}=\sqrt{3}{B}_0$