A positive charged particle with q=851 μC moves at 704m/s along the x-axis. It enters a region in which there is a magnetic field of magnitude 1.38T, directed at an angle of 61.5° with the x-axis and lying in the xy-plane. Find the magnitude of the magnetic force (unit in N) on the charged particle at this moment.
Added by Nathaniel W.
Step 1
To find the magnitude of the magnetic force on the charged particle, we can use the formula for the magnetic force on a charged particle moving in a magnetic field: \[ F = q \cdot v \cdot B \cdot \sin(\theta) \] where: - \( F \) is the magnetic force, - \( q \) Show more…
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