JEE MAIN - Physics (2020 - 2nd September Evening Slot - No. 9)
A wire carrying current I is bent in the shape
ABCDEFA as shown, where rectangle ABCDA
and ADEFA are perpendicular to each other. If
the sides of the rectangles are of lengths a and
b, then the magnitude and direction of
magnetic moment of the loop ABCDEFA is
_2nd_September_Evening_Slot_en_9_1.png)
$$\sqrt 2 $$abI, along $$\left( {{{\widehat j} \over {\sqrt 5 }} + {{2\widehat k} \over {\sqrt 5 }}} \right)$$
abI, along $$\left( {{{\widehat j} \over {\sqrt 5 }} + {{2\widehat k} \over {\sqrt 5 }}} \right)$$
$$\sqrt 2 $$abI, along $$\left( {{{\widehat j} \over {\sqrt 2 }} + {{\widehat k} \over {\sqrt 2 }}} \right)$$
abI, along $$\left( {{{\widehat j} \over {\sqrt 2 }} + {{\widehat k} \over {\sqrt 2 }}} \right)$$
Explanation
For ABCD
$${\overrightarrow M _1}$$ = abI$$\widehat k$$
For DEFA
$${\overrightarrow M _2}$$ = abI$$\widehat j$$
$$\overrightarrow M = {\overrightarrow M _1} + {\overrightarrow M _2}$$
= $$abI\left( {\widehat k + \widehat j} \right)$$
= $$abI\sqrt 2 \left( {{{\widehat k} \over {\sqrt 2 }} + {{\widehat j} \over {\sqrt 2 }}} \right)$$
$${\overrightarrow M _1}$$ = abI$$\widehat k$$
For DEFA
$${\overrightarrow M _2}$$ = abI$$\widehat j$$
$$\overrightarrow M = {\overrightarrow M _1} + {\overrightarrow M _2}$$
= $$abI\left( {\widehat k + \widehat j} \right)$$
= $$abI\sqrt 2 \left( {{{\widehat k} \over {\sqrt 2 }} + {{\widehat j} \over {\sqrt 2 }}} \right)$$
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