Verified Solution • 2025–26 Session moderate

QJEEM20Phd6dc: JEE Main 2020 Physics - Class 12 JEEM20Phd6dc Solution: A wire carrying current I is bent in the shape ABCDEFA as…

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 JEE Main 2020 (Online) 2nd September Evening Slot Physics - Magnetic Effect of Current Question 134 English

Multiple Choice Options

  • A 2\sqrt 2 abI, along (j^5+2k^5)\left( {{{\widehat j} \over {\sqrt 5 }} + {{2\widehat k} \over {\sqrt 5 }}} \right)
  • B abI, along (j^5+2k^5)\left( {{{\widehat j} \over {\sqrt 5 }} + {{2\widehat k} \over {\sqrt 5 }}} \right)
  • C 2\sqrt 2 abI, along (j^2+k^2)\left( {{{\widehat j} \over {\sqrt 2 }} + {{\widehat k} \over {\sqrt 2 }}} \right)
  • D abI, along (j^2+k^2)\left( {{{\widehat j} \over {\sqrt 2 }} + {{\widehat k} \over {\sqrt 2 }}} \right)

Step-by-Step Verified Solution

For ABCD

M→1{\overrightarrow M _1} = abI k^\widehat k

For DEFA

M→2{\overrightarrow M _2} = abI j^\widehat j

M→=M→1+M→2\overrightarrow M = {\overrightarrow M _1} + {\overrightarrow M _2}

= abI(k^+j^)abI\left( {\widehat k + \widehat j} \right)

= abI2(k^2+j^2)abI\sqrt 2 \left( {{{\widehat k} \over {\sqrt 2 }} + {{\widehat j} \over {\sqrt 2 }}} \right)

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring