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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Physics
Chapter
Magnetic Effects of Current

BITSAT 2019 · Physics · Magnetic Effects of Current

A proton is revolving on a circular path of radius 2 mm with...

A proton is revolving on a circular path of radius 22 mm with frequency 1010 Hz. Magnetic dipole moment associated with proton is

Options

  1. A2×10−24A−m22 \times {10}^{- 24}A - m^{2}
  2. B4×10−24A−m24 \times {10}^{- 24}A - m^{2}
  3. C3×10−20A−m23 \times {10}^{- 20}A - m^{2}
  4. D6×10−20A−m26 \times {10}^{- 20}A - m^{2}

Correct answer

A. 2×10−24A−m22 \times {10}^{- 24}A - m^{2}

Step-by-step solution

Given, charge on proton, q=eq = e
=1.6×10−19C= 1. 6 \times {10}^{- 19}C

Radius, r=2mm=2×10−3cmr = 2mm = 2 \times {10}^{- 3}cm

Frequency, f=10Hzf = 10Hz

Current associated by proton is given by

I=qtI = \dfrac{q}{t} & & & & & & & & & & [T→[ T \to time period ]]

=qf[∵T=1f]</p><p>= qf \left[∵ T = \dfrac{1}{f}\right]</p><p>= 1. 6 ×10−19×10=1.6×10−18\times {10}^{- 19} \times10 = 1. 6 \times {10}^{- 18} A</p><p></p><p>∴Magneticdipolemoment,Magnetic dipole moment,M = IA</p><p></p><p>= I ⋅\pir2=1.6×10−18×3.14×(2×10−3)2\cdot \pir^{2} =1. 6 \times {10}^{- 18} \times3. 14 \times \left(2 \times {10}^{- 3}\right)^{2}
=2×10−24A−m2= 2 \times {10}^{- 24}A - m^{2}

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