Question
A proton is accelerated from rest by a potential difference of $400 .$ V. The proton enters a uniform magnetic field and follows a circular path of radius $20.0 \mathrm{~cm} .$ Determine the magnitude of the magnetic field.
Step 1
First, we need to find the final velocity of the proton after being accelerated by the potential difference. We can use the following equation: qV = (1/2)mv^2 where q is the charge of the proton, V is the potential difference, m is the mass of the proton, and v Show more…
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A proton moves through a uniform electric field $\mathbf{E}=50.0 \mathbf{j} \mathrm{V} / \mathrm{m}$ and a uniform magnetic field $\mathbf{B}=$ $(0.200 \mathbf{i}+0.300 \mathbf{j}+0.400 \mathbf{k}) \mathrm{T}$. Determine the acceleration of the proton when it has a velocity $\mathbf{v}=200 \mathbf{i} \mathrm{m} / \mathrm{s}$.
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