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Physics and Mathematics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPhysics and Mathematics2 Marks
Question
The force on a charged particle due to electric and magnetic fields is given by $\overrightarrow{\text{F}}=\text{q}\overrightarrow{\text{E}}+\text{q}\overrightarrow{\text{v}}\times\overrightarrow{\text{B}}.$ Suppose $\overrightarrow{\text{E}}$ is along the X-axis and $\overrightarrow{\text{B}}$ along the Y-axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero?
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Answer

Given $\overrightarrow{\text{F}}=\text{q}\overrightarrow{\text{E}}+\text{q}(\overrightarrow{\text{v}}\times\overrightarrow{\text{B}})=0$$\Rightarrow\overrightarrow{\text{E}}=-(\overrightarrow{\text{v}}\times\overrightarrow{\text{B}})$
So, the direction of $\overrightarrow{\text{v}}\times\overrightarrow{\text{B}}$ should be opposite to the direction of $\overrightarrow{\text{E}}.$ Hence, $\overrightarrow{\text{v}}$ should be in the positive yz-plane. Again, $\text{E = vB}\sin\theta\Rightarrow\text{v}=\frac{\text{E}}{\text{B}\sin\theta}$ For v to be minimum, $\theta=90^{\circ}$ and so $\text{v}_{\text{min}}=\frac{\text{F}}{\text{B}}$ So, the particle must be projected at a minimum speed of E/B along +ve z-axis$(\theta=90^{\circ})$ as shown in the figure, so that the force is zero.

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