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JEE Main 201511 AprilOnline
Exam
JEE Main
Year
2015
Date
11 April
Mode
Online
Subject
Physics
Chapter
Magnetic Properties of Matter

JEE Main 2015 · 11 April · Online · Physics · Magnetic Properties of Matter

A short bar magnet is placed in the magnetic meridian of the earth...

A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are found at a distance of 30cm30cm from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in Am2{Am}^{2} is close to:

(Given μ04π=10−7\dfrac{{\mu}_{0}}{4\pi} = {10}^{- 7} in SI units and BH={B}_{H} = Horizontal component of earth's magnetic field =3.6×10−5= 3.6 \times {10}^{- 5} Tesla.)

A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are

Options

  1. A4.9
  2. B14.6
  3. C19.4
  4. D9.7

Correct answer

D. 9.7

Step-by-step solution

Horizontal component of earth's magnetic field is along geographical south its North.

M→=\overset{\to}{M} = Magnetic dipole moment.

A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are

At neutral point on the equatorial line the magnetic field B′{B}^{′} cancels BH{B}_{H}

∴B′=BH∴ {B}^{′} = {B}_{H}

⇒μ04πMx3=BH⇒ \dfrac{{\mu}_{0}}{4\pi}\dfrac{M}{{x}^{3}} = {B}_{H}

⇒10−7×M(30100)3=3.6×10−5⇒ \dfrac{{10}^{- 7} \times M}{{\left(\dfrac{30}{100}\right)}^{3}} = 3.6 \times {10}^{- 5}

⇒10−7×103×M27=3.6×10−5⇒ \dfrac{{10}^{- 7} \times {10}^{3} \times M}{27} = 3.6 \times {10}^{- 5}

M=27×3.6×10−510−4M = \dfrac{27 \times 3.6 \times {10}^{- 5}}{{10}^{- 4}}

M=27×3.6×10−1M = 27 \times 3.6 \times {10}^{- 1}

=9.72= 9.72

M≈9.7M \approx 9.7
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