An electron is moving with a velocity (2 \overline{i}+3 \overline{j}) {\ms}^{-1} in an electric field (3 \overline{i}+6 \overline{j}+2 \overline{k}) {\Vs}^{-1} and a magnetic field of (2 \overline{j}+3 \overline{k}) {\T}. Then the magnitude and direction (with x-axis) of the Lorentz force acting on the electron is [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : 2.15×10−18N,θ=cos−1(2√5)

9.6 \times 10^{-19} {\N}, \theta=\cos ^{-1} ( {2}/{\sqrt{5}})

9.6 \times 10^{-19} {\N}, \theta=\cos ^{-1} ( {5}/{\sqrt{2}})

2.15 \times 10^{-18} {\N}, \theta=\cos ^{-1} ( {2}/{\sqrt{5}})

2.15 \times 10^{-18} {\N}, \theta=\cos ^{-1} ( {5}/{\sqrt{2}})

Moving Charges and Magnetism

Section: Physics

Question : 32 of 32

Marks: +1, -0

An electron is moving with a velocity (2i+3j)ms−1 in an electric field (3i+6j+2k)Vs−1 and a magnetic field of (2j+3k)T. Then the magnitude and direction (with x-axis) of the Lorentz force acting on the electron is

[AP EAPCET 18 May 2024 Shift 1]

9.6×10−19N,‌‌θ=cos−1(‌
2
√5
)
9.6×10−19N,‌‌θ=cos−1(‌
5
√2
)
2.15×10−18N,‌‌θ=cos−1(‌
2
√5
)
2.15×10−18N,‌‌θ=cos−1(‌
5
√2
)

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