Question

Three current carrying elements having same length carries current $i,2i$ and $4i$ respectively. Magnetic fields due to wires be $B_1,B_2$ and $B_3$ respectively at a point $\text{P}$ whose distance from the current elements is $r,2r$ and $4r$ respectively. Then, $B_1:B_2:B_3$ is

• A. $1:2:4$
• B. $2:1:4$
• C. $4:2:1$
• D. $4:1:2$

Answer 1

The correct option is C $4:2:1$

According to Biot-Savart law,

Magnetic field due to a current element is given by,

$dB=\frac{{\mu }_0}{4\pi }\frac{idl\sin \theta }{r^2}$

So, $B_1:B_2:B_3$

$\Rightarrow B_1:B_2:B_3=\frac{{\mu }_0}{4\pi }\frac{i_{1}dl\sin \theta }{r_1^2}:\frac{{\mu }_0}{4\pi }\frac{i_{2}dl\sin \theta }{r_2^2}:\frac{{\mu }_0}{4\pi }\frac{i_{3}dl\sin \theta }{r_3^2}$

$\Rightarrow B_1:B_2:B_3=\frac{i_1}{r_1^2}:\frac{i_2}{r_2^2}:\frac{i_3}{r_3^2}$

$\Rightarrow B_1:B_2:B_3=\frac{i}{r^2}:\frac{2i}{(2r{)}^2}:\frac{4i}{(4r{)}^2}$

$\Rightarrow B_1:B_2:B_3=1:\frac{1}{2}:\frac{1}{4}$

$\therefore B_1:B_2:B_3=4:2:1$