QuestionJuly 2, 2025

An electron has a velocity of 4.5times 10^4m/s perpendicular to a magnetic field of 3 .5 mT. Determine the magnetic force acting on the electron. 2.50times 10^-17N 2.50times 10^-17N 3.65times 10^-16N 3.65times 10^-16N

An electron has a velocity of 4.5times 10^4m/s perpendicular to a magnetic field of 3 .5 mT. Determine the magnetic force acting on the electron. 2.50times 10^-17N 2.50times 10^-17N 3.65times 10^-16N 3.65times 10^-16N
An electron has a velocity of 4.5times 10^4m/s perpendicular to a magnetic field of 3 .5 mT. Determine the magnetic force acting on the electron. 2.50times 10^-17N 2.50times 10^-17N 3.65times 10^-16N 3.65times 10^-16N

Solution
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Answer

\( 2.52 \times 10^{-17} \mathrm{~N} \) Explanation 1. Identify the formula for magnetic force The magnetic force on a charged particle is given by **F = qvB\sin(\theta)**. Here, q is the charge of the electron, v is the velocity, B is the magnetic field strength, and \theta is the angle between the velocity and the magnetic field. 2. Substitute known values For an electron, q = -1.6 \times 10^{-19} \mathrm{~C}, v = 4.5 \times 10^{4} \mathrm{~m/s}, B = 3.5 \times 10^{-3} \mathrm{~T}, and \theta = 90^\circ. Since \sin(90^\circ) = 1, the formula simplifies to F = qvB. 3. Calculate the magnetic force F = (-1.6 \times 10^{-19} \mathrm{~C})(4.5 \times 10^{4} \mathrm{~m/s})(3.5 \times 10^{-3} \mathrm{~T}) = 2.52 \times 10^{-17} \mathrm{~N} (ignoring the negative sign as force magnitude is positive).

Explanation

1. Identify the formula for magnetic force<br /> The magnetic force on a charged particle is given by **$F = qvB\sin(\theta)$**. Here, $q$ is the charge of the electron, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field.<br /><br />2. Substitute known values<br /> For an electron, $q = -1.6 \times 10^{-19} \mathrm{~C}$, $v = 4.5 \times 10^{4} \mathrm{~m/s}$, $B = 3.5 \times 10^{-3} \mathrm{~T}$, and $\theta = 90^\circ$. Since $\sin(90^\circ) = 1$, the formula simplifies to $F = qvB$.<br /><br />3. Calculate the magnetic force<br /> $F = (-1.6 \times 10^{-19} \mathrm{~C})(4.5 \times 10^{4} \mathrm{~m/s})(3.5 \times 10^{-3} \mathrm{~T}) = 2.52 \times 10^{-17} \mathrm{~N}$ (ignoring the negative sign as force magnitude is positive).
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