Question

Two straight long conductors AOB and COD are perpendicular to each other and carry currents $ {i_1} $ and $ {i_2} $ .the magnitude of the magnetic induction at a point P at a distance $ a $ from the point O in a direction perpendicular to the plane ABCD is?
(A) $ \dfrac{{{\mu _0}}}{{2\pi a}}\left( {{i_1} + {i_2}} \right) $
(B) $ \dfrac{{{\mu _0}}}{{2\pi a}}\left( {{i_1} - {i_2}} \right) $
(C) $ \dfrac{{{\mu _0}}}{{2\pi a}}{(i_1^2 + i_2^2)^{\dfrac{1}{2}}} $
(D) $ \dfrac{{{\mu _0}}}{{2\pi a}}\left( {\dfrac{{{i_1}{i_2}}}{{{i_1} + {i_2}}}} \right) $

Answer 1

Hint :Find the magnetic field due to each wire at P and calculate the magnitude of the net magnetic field there. From Biot- Savart law the magnetic field at perpendicular distance $ a $ from a wire is given as, $ \dfrac{{{\mu _0}i}}{{2\pi a}} $ . Where, $ {\mu _0} $ is the absolute magnetic permeability , $ i $ is the current flowing through it. The direction can be found from the right hand palm rule.

Complete Step By Step Answer:
We know from Biot- Savart law that magnetic field at a distance $ a $ due to wire of infinite length is given by, $ \dfrac{{{\mu _0}i}}{{2\pi a}} $ . Where, $ {\mu _0} $ is the absolute magnetic permeability, $ i $ is the current flowing through it. The direction can be found from the right hand palm rule.
Then we have the magnetic field due to the first wire with current $ {i_1} $ at the point P is, $ {B_1} = \dfrac{{{\mu _0}{i_1}}}{{2\pi a}} $ .
If we take the direction of current on it along the positive X -axis then The direction of the magnetic field will be along the negative Y-axis. Hence, with vector notation we can write it as, $ {\vec B_1} = \dfrac{{{\mu _0}{i_1}}}{{2\pi a}}( - \hat j) $
We can find the magnetic field due to the wire COD as, $ {B_2} = \dfrac{{{\mu _0}{i_2}}}{{2\pi a}} $ .
If we take the direction of current along positive Y-axis then the direction of current will be along positive X-axis. Hence, with vector notation we can write it as, $ {\vec B_2} = \dfrac{{{\mu _0}{i_2}}}{{2\pi a}}\hat i $
Hence, Net magnetic field at $ a $ can be written as ,
 $ {\vec B_{net}} = {\vec B_1} + {\vec B_2} $
 Putting the values we get,
 $ {\vec B_{net}} = - \dfrac{{{\mu _0}{i_1}}}{{2\pi a}}\hat j + \dfrac{{{\mu _0}{i_2}}}{{2\pi a}}\hat i $
Therefore, if we take the magnitude of it that becomes,
 $ \left| {{{\vec B}_{net}}} \right| = \sqrt {{{(\dfrac{{{\mu _0}{i_2}}}{{2\pi a}})}^2} + {{(\dfrac{{{\mu _0}{i_1}}}{{2\pi a}})}^2}} $
On simplifying we get,
 $ \left| {{{\vec B}_{net}}} \right| = \dfrac{{{\mu _0}}}{{2\pi a}}\sqrt {i_1^2 + i_2^2} $
Hence, this is our answer.
Hence, Option ( C) is correct.

Note :
Magnetic fields due to two wires kept perpendicular cannot be zero, since the direction of them is always along perpendicular axes. But, the magnetic field due to two wires kept parallel can be zero. The magnetic field will be zero at the middle of the wires if the direction of current is the same for both the wires.