JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 19)
There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is 1 : 1. The magnetic field at point P is ____________.
_1st_September_Evening_Shift_en_19_1.png)
$${{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]$$
$${{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} - (x + y)} \right]$$
$${{{\mu _0}Ixy} \over {4\pi }}\left[ {\sqrt {{x^2} + {y^2}} - (x + y)} \right]$$
$${{{\mu _0}Ixy} \over {4\pi }}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]$$
Explanation
_1st_September_Evening_Shift_en_19_2.png)
Bdue to wire (1) = $${{{\mu _0}I} \over {4\pi y}}\left[ {\sin 90 + \sin {\theta _1}} \right]$$
$$ = {{{\mu _0}} \over {4\pi }}{I \over y}\left( {1 + {x \over {\sqrt {{x^2} + {y^2}} }}} \right)$$ ....(1)
Bdue to wire (2) = $${{{\mu _0}} \over {4\pi }}{1 \over x}(\sin 90^\circ + \sin {\theta _2})$$
$$ = {{{\mu _0}} \over {4\pi }}{1 \over x}\left( {1 + {y \over {\sqrt {{x^2} + {y^2}} }}} \right)$$ ....(2)
Total magnetic field
$$B = {B_1} + {B_2}$$
$$B = {{{\mu _0}I} \over {4\pi }}\left[ {{1 \over y} + {x \over {y\sqrt {{x^2} + {y^2}} }} + {1 \over x} + {y \over {x\sqrt {{x^2} + {y^2}} }}} \right]$$
$$B = {{{\mu _0}I} \over {4\pi }}\left[ {{{x + y} \over {xy}} + {{{x^2} + {y^2}} \over {xy\sqrt {{x^2} + {y^2}} }}} \right]$$
$$B = {{{\mu _0}I} \over {4\pi }}\left[ {{{x + y} \over {xy}} + {{\sqrt {{x^2} + {y^2}} } \over {xy}}} \right]$$
$$B = {{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]$$
Option (1)
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