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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Magnetic Field due to a Current

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Magnetic Field due to a Current

An electron completes 3 × 10^5 revolutions per second in a circle...

An electron completes 3×1053 \times 10^5 revolutions per second in a circle of radius 0.50.5 angstrom. Determine the magnetic field BB at the centre of the circle.

Options

  1. A6.0×10−10 T6.0 \times 10^{-10} \text{ T}
  2. B3.0×10−10 T3.0 \times 10^{-10} \text{ T}
  3. C4.8×10−14 T4.8 \times 10^{-14} \text{ T}
  4. D1.2×10−9 T1.2 \times 10^{-9} \text{ T}

Correct answer

A. 6.0×10−10 T6.0 \times 10^{-10} \text{ T}

Step-by-step solution

The equivalent current II due to the revolving electron is given by:

I=e×fI = e \times f

Substituting the given values, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C} and f=3×105 Hzf = 3 \times 10^5 \text{ Hz}:

I=1.6×10−19×3×105=4.8×10−14 AI = 1.6 \times 10^{-19} \times 3 \times 10^5 = 4.8 \times 10^{-14} \text{ A}

The magnetic field BB at the centre of the circular path is:

B=μ0I2rB = \dfrac{\mu_0 I}{2r}

Substituting μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7} \text{ T m/A} and r=0.5×10−10 mr = 0.5 \times 10^{-10} \text{ m}:

B=4π×10−7×4.8×10−142×0.5×10−10B = \dfrac{4\pi \times 10^{-7} \times 4.8 \times 10^{-14}}{2 \times 0.5 \times 10^{-10}}

B=19.2π×10−2110−10B = \dfrac{19.2\pi \times 10^{-21}}{10^{-10}}

B=19.2π×10−11 TB = 19.2\pi \times 10^{-11} \text{ T}

B≈60.3×10−11 T≈6.0×10−10 TB \approx 60.3 \times 10^{-11} \text{ T} \approx 6.0 \times 10^{-10} \text{ T}
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