JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 4)
Two very long, straight, and insulated wires are
kept at 90° angle from each other in xy-plane
as shown in the figure. These wires carry
currents of equal magnitude I, whose directions
are shown in the figure. The net magnetic field
at point P will be :
_8th_April_Evening_Slot_en_4_1.png)
$${{ + {\mu _0}I} \over {\pi d}}\left( {\mathop z\limits^ \wedge } \right)$$
$$ - {{{\mu _0}I} \over {2\pi d}}\left( {\mathop x\limits^ \wedge + \mathop y\limits^ \wedge } \right)$$
Zero
$$ {{{\mu _0}I} \over {2\pi d}}\left( {\mathop x\limits^ \wedge + \mathop y\limits^ \wedge } \right)$$
Explanation
Magnetic field at point P
$${\overrightarrow B _{net}} = {{{\mu _0}i} \over {2\pi d}}( - \widehat k) + {{{\mu _0}i} \over {2\pi d}}(\widehat k) = 0$$
$${\overrightarrow B _{net}} = {{{\mu _0}i} \over {2\pi d}}( - \widehat k) + {{{\mu _0}i} \over {2\pi d}}(\widehat k) = 0$$
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