Question
Two identical long conducting wires $$AOB$$ and $$COD$$ are placed at right angle to each other, with one above other such that $$'O'$$ is their common point for the two. The wires carry $${I_1}$$ and $${I_2}$$ currents respectively. Point $$'P'$$ is lying at distance $$'d'$$ from $$'O'$$ along a direction perpendicular to the plane containing the wires. The magnetic field at the point $$'P'$$ will be:
A.
$$\frac{{{\mu _0}}}{{2\pi d}}\left( {\frac{{{I_1}}}{{{I_2}}}} \right)$$
B.
$$\frac{{{\mu _0}}}{{2\pi d}}\left( {{I_1} + {I_2}} \right)$$
C.
$$\frac{{{\mu _0}}}{{2\pi d}}\left( {I_1^2 - I_2^2} \right)$$
D.
$$\frac{{{\mu _0}}}{{2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{\frac{1}{2}}}$$
Answer :
$$\frac{{{\mu _0}}}{{2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{\frac{1}{2}}}$$
Solution :
Net magnetic field, $$B = \sqrt {B_1^2 + B_2^2} $$
$$\eqalign{
& = \sqrt {{{\left( {\frac{{{\mu _0}{I_1}}}{{2\pi d}}} \right)}^2} + {{\left( {\frac{{{\mu _0}{I_2}}}{{2\pi d}}} \right)}^2}} \cr
& = \frac{{{\mu _0}}}{{2\pi d}}\sqrt {I_1^2 + I_2^2} . \cr} $$