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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Electromagnetic Waves

JIPMER 2016 · Physics · Electromagnetic Waves

A beam of light travelling along X -axis is described by the...

A beam of light travelling along XX-axis is described by the electric field EY=600VmE_Y=600 \frac{\mathrm{V}}{\mathrm{m}}. sin⁡ω(t−xc)\sin \omega\left(t-\frac{x}{c}\right), the maximum magnetic force on a charge q=2eq=2 e, moving along YY-axis with the speed of 3×108 m/s3 \times 10^8 \mathrm{~m} / \mathrm{s} is

Options

  1. A19.2×10−17 N19.2 \times 10^{-17} \mathrm{~N}
  2. B1.92×10−17 N1.92 \times 10^{-17} \mathrm{~N}
  3. C0.192 N
  4. DNone of these

Correct answer

B. 1.92×10−17 N1.92 \times 10^{-17} \mathrm{~N}

Step-by-step solution

Given, electric field, Ey=600Vm⋅sin⁡ω(t−xc)...(i)Weknowthat,E_y=600 \frac{\mathrm{V}}{\mathrm{m}} \cdot \sin \omega\left(t-\frac{x}{\mathrm{c}}\right)...(i) We know that,E=E_0 \sin \omega\left(t-\frac{x}{c}\right).....(ii)OncomparingEqs.(i)and(ii),weget.....(ii) On comparing Eqs. (i) and (ii), we getE_0=600 \mathrm{~V} / \mathrm{m} and C=3 \times 10^8 \mathrm{~m} / \mathrm{s}Maximummagneticfield,Maximum magnetic field,B0=E0CB0=6003×108=2×10−6 T\begin{aligned} & B_0=\frac{E_0}{C} \\ & B_0=\frac{600}{3 \times 10^8}=2 \times 10^{-6} \mathrm{~T}\end{aligned}$\begin{aligned} \therefore \quad F_m & =q v B_0=2 \mathrm{ev} B_0 \\ & =2 \times 1.6 \times 10^{-19} \times 3 \times 10^8 \times 2 \times 10^{-6} \\ & =19.2 \times 10^{-17} \mathrm{~N}\end{aligned}
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