Question 245: Magnetism and Matter

Question

Two identical conducting wires $AOB$ and $COD$ are placed at right angles to each other. The wire $AOB$ carries an electric current ${I_1}$ and $COD$ carries a current ${I_2}$. The magnetic field on a point lying at a distance $d$ from $O$, in a direction perpendicular to the plane of the wires $AOB$ and $COD$ , will be given by

Accepted Answer

Clearly, the magnetic fields at a point $P,$ equidistant from $AOB$ and $COD$ will have directions perpendicular to each  other, as they are placed normal to each other.
$	herefore$ Resultant field, $B = \sqrt {B_1^2 + B_2^2} $
But ${B_1} = {{{\mu _0}{I_1}} \over {2\pi d}}$ and ${B_2} = {{{\mu _0}{I_2}} \over {2\pi d}}$
$	herefore$ $B = \sqrt {{{\left( {{{{\mu _0}} \over {2\pi d}}} ight)}^2}\left( {I_1^2 + I_2^2} ight)} $
or, $B = {{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} ight)^{1/2}}$

Answer

${{{\mu _0}} \over {2\pi d}}\left( {I_1^2 + I_2^2} \right)$

Answer

${{{\mu _0}} \over {2\pi }}{\left( {{{{I_1} + {I_2}} \over d}} \right)^{{1 \over 2}}}$

Answer

${{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{{1 \over 2}}}$

Answer

${{{\mu _0}} \over {2\pi d}}\left( {{I_1} + {I_2}} \right)$

Magnetism and Matter

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