JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 22)
An electron moves through a uniform magnetic field $$\vec{B}=B_0 \hat{i}+2 B_0 \hat{j} T$$. At a particular instant of time, the velocity of electron is $$\vec{u}=3 \hat{i}+5 \hat{j} \mathrm{~m} / \mathrm{s}$$. If the magnetic force acting on electron is $$\vec{F}=5 e \hat{k} N$$, where $$e$$ is the charge of electron, then the value of $$B_0$$ is _________ $$T$$.
Answer
5
Explanation
$$\begin{aligned}
& \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \\
& 5 \mathrm{e} \hat{\mathrm{k}}=\mathrm{e}(3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}) \times\left(\mathrm{B}_0 \hat{\mathrm{i}}+2 \mathrm{~B}_0 \hat{\mathrm{j}}\right) \\
& 5 \mathrm{e} \hat{\mathrm{k}}=\mathrm{e}\left(6 \mathrm{~B}_0 \hat{\mathrm{k}}-5 \mathrm{~B}_0 \hat{\mathrm{k}}\right) \\
& \Rightarrow \mathrm{B}_0=5 \mathrm{~T}
\end{aligned}$$
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