AP EAMCET · PHYSICS · Magnetic Effects of Current
An electron is moving with a velocity \((2 \bar{i}+3 \bar{j}) \mathrm{ms}^{-1}\) in an electric field \((3 \bar{i}+6 \bar{j}+2 \bar{k}) \mathrm{Vs}^{-1}\) and a magnetic field of \((2 \bar{j}+3 \bar{k}) \mathrm{T}\). Then the magnitude and direction (with \(x\)-axis) of the Lorentz force acting on the electron is
- A \(9.6 \times 10^{-19} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
- B \(9.6 \times 10^{-19} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{5}{\sqrt{2}}\right)\)
- C \(2.15 \times 10^{-18} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
- D \(2.15 \times 10^{-18} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{5}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(2.15 \times 10^{-18} \mathrm{~N}, \theta=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
Step-by-step Solution
Detailed explanation
\(v=(2 \hat{i}+3 \hat{j}) \mathrm{m} / \mathrm{s}, E=(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \mathrm{Vs}\) \(\overrightarrow{\mathrm{B}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}) \mathrm{T}\) The lorentz force acting on electron is…
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