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A electron experiences a force $\left( {4.0\,\hat i + 3.0\,\hat j} \right)\times 10^{-13} N$ in a uniform magnetic field when its velocity is $2.5\,\hat k \times \,{10^7} ms^{-1}$. When the velocity is redirected and becomes $\left( {1.5\,\hat i - 2.0\,\hat j} \right) \times {10^7}$, the magnetic force of the electron is zero. The magnetic field $\vec B$ is :
$-0.075\,\hat i + 0.1\,\hat j$
$0.1\,\hat i + 0.075\,\hat j$
$0.075\,\hat i - 0.1\,\hat j + \hat k$
$0.075\,\hat i - 0.1\,\hat j$
Solution
$\overrightarrow{F_{m}}=q(\vec{V} \times \vec{B})=-e(\vec{V} \times \vec{B})=e(\vec{B} \times \vec{V})$
$(4 \hat{i} \times 3 \hat{j}) \times 10^{-13}$
$=16 \times 10^{-19}\left(B_{x} \hat{i}+B_{y} \hat{j}\right) \times 2.5 \times 10^{7} \hat{k}$
$(4 \hat{i}+3 \hat{j})=\left(40 B_{y} \hat{i}-40 B_{x} \hat{j}\right)$
$\vec{B}=-0.075 \hat{i}+0.1 \hat{j}$
