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Magnetics question

2022 · 26 Jul · Shift 2 · Q54
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  5. /2022 · 26 Jul · Shift 2 · Q54

Magnetics question

2022 · 26 Jul · Shift 2 · Q54

JEE MainPhysicsMagneticsMCQ+4 / −1
Two concentric circular loops of radii r1=30 cmr_{1}=30 \mathrm{~cm}r1​=30 cm and r2=50 cmr_{2}=50 \mathrm{~cm}r2​=50 cm are placed in X−Y\mathrm{X}-\mathrm{Y}X−Y plane as shown in the figure. A current I=7 AI=7 \mathrm{~A}I=7 A is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately : JEE Main 2022 (Online) 26th July Evening Shift Physics - Magnetic Effect of Current Question 92 English
  1. A
    72k^ Am2\frac{7}{2} \hat{k} \,\mathrm{Am}^{2}27​k^Am2
  2. B
    −72k^ Am2-\frac{7}{2} \hat{k} \,\mathrm{Am}^{2}−27​k^Am2
  3. C
    7 k^ Am2{7}\, \hat{k} \,\mathrm{Am}^{2}7k^Am2
  4. D
    −7 k^ Am2{-7}\, \hat{k} \,\mathrm{Am}^{2}−7k^Am2
View written solutionFree

Correct answer: B

  1. Magnetic moment of a current loop

For a circular current loop,

m⃗=I A⃗\vec m = I\,\vec Am=IA

where A⃗\vec AA is the area vector of magnitude πr2\pi r^2πr2 and direction is given by the right-hand rule.

So,

m⃗=Iπr2 n^\vec m = I\pi r^2\,\hat nm=Iπr2n^
  1. Direction of magnetic moment

The two loops are concentric and lie in the XYXYXY-plane, so their magnetic moments must be along either +k^+\hat k+k^ or −k^-\hat k−k^.

From the figure (as implied by the options and standard convention), the current directions in the two loops are opposite, so their magnetic moments are opposite.

  • Outer loop contributes along one of ±k^\pm \hat k±k^
  • Inner loop contributes along the opposite direction

Using the shown current direction, the net magnetic moment is along −k^-\hat k−k^.

  1. Magnitude of net magnetic moment

Since directions are opposite,

mnet=Iπ(r22−r12)m_{\text{net}} = I\pi (r_2^2-r_1^2)mnet​=Iπ(r22​−r12​)

with

r1=30 cm=0.3 m,r2=50 cm=0.5 mr_1 = 30\text{ cm} = 0.3\text{ m}, \qquad r_2 = 50\text{ cm} = 0.5\text{ m}r1​=30 cm=0.3 m,r2​=50 cm=0.5 m

Hence,

r22−r12=(0.5)2−(0.3)2=0.25−0.09=0.16r_2^2-r_1^2 = (0.5)^2-(0.3)^2 = 0.25-0.09 = 0.16r22​−r12​=(0.5)2−(0.3)2=0.25−0.09=0.16

Therefore,

mnet=7π(0.16)=1.12πm_{\text{net}} = 7\pi(0.16) = 1.12\pimnet​=7π(0.16)=1.12π

Using π≈227\pi \approx \frac{22}{7}π≈722​,

mnet≈1.12×227=3.52≈3.5m_{\text{net}} \approx 1.12 \times \frac{22}{7} = 3.52 \approx 3.5mnet​≈1.12×722​=3.52≈3.5

Thus,

m⃗net≈−3.5 k^ A m2\vec m_{\text{net}} \approx -3.5\,\hat k\,\text{A m}^2mnet​≈−3.5k^A m2
  1. Match with options
−3.5 k^ A m2=−72k^ A m2-3.5\,\hat k\,\text{A m}^2 = -\frac{7}{2}\hat k\,\text{A m}^2−3.5k^A m2=−27​k^A m2

So the correct option is:

B. −72k^ Am2-\frac{7}{2}\hat k\,\mathrm{A m^2}−27​k^Am2

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