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Magnetics question

2020 · 2 Sep · Shift 2 · Q51
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Magnetics question

2020 · 2 Sep · Shift 2 · Q51

JEE MainPhysicsMagneticsMCQ+4 / −1
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 144 English
  1. A
    2\sqrt 22​ abI, along (j^5+2k^5)\left( {{{\widehat j} \over {\sqrt 5 }} + {{2\widehat k} \over {\sqrt 5 }}} \right)(5​j​​+5​2k​)
  2. B
    abI, along (j^5+2k^5)\left( {{{\widehat j} \over {\sqrt 5 }} + {{2\widehat k} \over {\sqrt 5 }}} \right)(5​j​​+5​2k​)
  3. C
    2\sqrt 22​ abI, along (j^2+k^2)\left( {{{\widehat j} \over {\sqrt 2 }} + {{\widehat k} \over {\sqrt 2 }}} \right)(2​j​​+2​k​)
  4. D
    abI, along (j^2+k^2)\left( {{{\widehat j} \over {\sqrt 2 }} + {{\widehat k} \over {\sqrt 2 }}} \right)(2​j​​+2​k​)
View written solutionFree

Correct answer: C

  1. Magnetic moment of a planar current loop

For any current loop,

m⃗=I A⃗\vec m = I\,\vec Am=IA

where A⃗\vec AA is the vector area of the loop.

If a loop is made of multiple planar parts, then the total magnetic moment is the vector sum of the moments of those parts.

  1. Interpret the bent loop

The wire is bent along A→B→C→D→E→F→AA\to B\to C\to D\to E\to F\to AA→B→C→D→E→F→A. It consists of two rectangular parts sharing the side ADADAD:

  • rectangle ABCDAABCDAABCDA
  • rectangle ADEFAADEFAADEFA

These two rectangles are perpendicular to each other.

Each rectangle has sides aaa and bbb, so area of each rectangle is

A=ab.A = ab.A=ab.

Hence each rectangular part contributes magnetic moment of magnitude

m0=Iab.m_0 = Iab.m0​=Iab.
  1. Direction of magnetic moment of each rectangle

From the options, the two rectangle area vectors must lie along mutually perpendicular directions j^\hat jj^​ and k^\hat kk^.

Using the right-hand rule according to the current direction shown in the figure, the magnetic moments of the two rectangles are along:

  • for rectangle ABCDAABCDAABCDA: Iab j^Iab\,\hat jIabj^​
  • for rectangle ADEFAADEFAADEFA: Iab k^Iab\,\hat kIabk^

Therefore,

m⃗=Iab j^+Iab k^=Iab(j^+k^).\vec m = Iab\,\hat j + Iab\,\hat k = Iab(\hat j + \hat k).m=Iabj^​+Iabk^=Iab(j^​+k^).
  1. Magnitude of resultant magnetic moment
∣m⃗∣=Iab12+12=2 abI.|\vec m| = Iab\sqrt{1^2+1^2} = \sqrt{2}\,abI.∣m∣=Iab12+12​=2​abI.
  1. Direction of resultant magnetic moment

The unit vector along m⃗\vec mm is

m^=j^+k^2.\hat m = \frac{\hat j + \hat k}{\sqrt{2}}.m^=2​j^​+k^​.

So the magnetic moment is

m⃗=2 abI(j^2+k^2).\boxed{\vec m = \sqrt{2}\,abI\left(\frac{\hat j}{\sqrt{2}}+\frac{\hat k}{\sqrt{2}}\right)}.m=2​abI(2​j^​​+2​k^​)​.
  1. Match with options

This corresponds to:

Option C\boxed{\text{Option C}}Option C​
  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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