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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Magnetic Properties of Matter

JEE Main · Physics · Magnetic Properties of Matter

A bar magnet of magnetic moment M is placed at a distance D with...

A bar magnet of magnetic moment MM is placed at a distance DD with its axis along positive XX-axis. Likewise, second bar magnet of magnetic moment MM is placed at a distance 2D2D on positive YY-axis and perpendicular to it as shown in the figure. The magnitude of magnetic field at the origin is \left|\overset{→}{B}\right| = \alpha\left[\dfrac{\mu0}{4\pi}\dfrac{M}{D3}\right].Thevalueof. The value of\alphamustbe(Assumemust be (AssumeD > >wherewherelisthelengthofmagnets)</p><p><imgsrc="/api/proxy−image?url=httpsis the length of magnets)</p><p><img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fquizrr%252Fquestion-assets%252Fts_eamcet%252Fpy35bdsq%252F55cc1234-8d90-412a-96f6-04932bad7503-image.png%3Falt%3Dmedia&clean=1&v=qxfig138" style="width: 287px; height: 267px;" / data-qx-orig-src="/assets/diagrams/qx-book-203cbee53a6c5a1a.png" data-qx-storage-src="/assets/diagrams/qx-book-203cbee53a6c5a1a.png"></p>

Options

  1. A22
  2. B158\dfrac{15}{8}
  3. C178\dfrac{17}{8}
  4. D98\dfrac{9}{8}

Correct answer

B. 158\dfrac{15}{8}

Step-by-step solution

Baxis=2μ0M4\piD3 and Bequator=μ0M8\piD3{B}_{axis} = \dfrac{2{μ}_{0}M}{4\piD^{3}} \text{ and } {B}_{equator} = \dfrac{{μ}_{0}M}{8\piD^{3}}

since~d=2Dd = 2D

{B}_{net} = {B}_{axis} - {B}_{equator} = \dfrac{{μ}_{0}M}{4\piD^{3}} \left(2 - \dfrac{1}{8}\right)</p><p>∴ {B}_{net} = 158μ0M4\piD3\dfrac{15}{8}\dfrac{{μ}_{0}M}{4\piD^{3}}⇒ \alpha = \dfrac{15}{8}</p></p>

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