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A charge $q$ enters a region having electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ with velocity $\mathbf{v}$. If it continues to move with the same velocity, then which of the following statement is not true
Options:
Solution:
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Verified Answer
The correct answer is:
$v \times E=B$
Lorentz force on electron,
$$
F=q(\mathbf{E}+\mathbf{v}+\mathbf{B})
$$
For regular motion, $F=0$
$$
\Rightarrow \quad \mathbf{E}+(\mathbf{v} \times \mathbf{B})=0
$$
$\Rightarrow \mathbf{E} \|(\mathbf{v} \times \mathbf{B})$, also $[\mathbf{v} \times \mathbf{B}$ is perpendicular to both $v$ and $\mathbf{B}]$
Now, $\mathbf{E} \perp \mathbf{v}$ and $\mathbf{E} \perp \mathbf{B}$ [(i) and (ii) are right]
$$
\Rightarrow \quad \mathbf{E} \cdot \mathbf{v}=0 \text { and } \mathbf{E} \cdot \mathbf{B}=0
$$
Now, if $\quad \mathbf{v} \cdot \mathbf{B}=0 \Rightarrow \mathbf{v} \perp \mathbf{B}$
So, $\mathbf{E} \perp \mathbf{v} \perp \mathbf{B}$
$$
\begin{aligned}
& \mathbf{E}=\mathbf{v} \times \mathbf{B} \\
& \mathbf{E} \times \mathbf{B}=(\mathbf{V} \times \mathbf{B}) \times \mathbf{B} \\
& v \times \mathbf{B}=v B \hat{\mathbf{n}} \\
& \Rightarrow \quad(\mathbf{v} B) \times \mathbf{B}=\mathbf{v}(\mathbf{B} \cdot \mathbf{B}) \\
& \Rightarrow \quad \mathbf{v}=\frac{\mathbf{E} \times \mathbf{B}}{\mathbf{B} \cdot \mathbf{B}} \\
& \mathbf{v} \times \mathbf{E}=\mathbf{v} \times \mathbf{v} \times \mathbf{B}=0 \neq \mathbf{B} \\
&
\end{aligned}
$$
[iii) is right]
Now, $\quad \mathbf{E}=\mathbf{v} \times \mathbf{B}$
$$
\mathbf{v} \times \mathbf{E}=\mathbf{v} \times \mathbf{v} \times \mathbf{B}=0 \neq \mathbf{B}
$$
So, option (d) is wrong.
$$
F=q(\mathbf{E}+\mathbf{v}+\mathbf{B})
$$
For regular motion, $F=0$
$$
\Rightarrow \quad \mathbf{E}+(\mathbf{v} \times \mathbf{B})=0
$$
$\Rightarrow \mathbf{E} \|(\mathbf{v} \times \mathbf{B})$, also $[\mathbf{v} \times \mathbf{B}$ is perpendicular to both $v$ and $\mathbf{B}]$
Now, $\mathbf{E} \perp \mathbf{v}$ and $\mathbf{E} \perp \mathbf{B}$ [(i) and (ii) are right]
$$
\Rightarrow \quad \mathbf{E} \cdot \mathbf{v}=0 \text { and } \mathbf{E} \cdot \mathbf{B}=0
$$
Now, if $\quad \mathbf{v} \cdot \mathbf{B}=0 \Rightarrow \mathbf{v} \perp \mathbf{B}$
So, $\mathbf{E} \perp \mathbf{v} \perp \mathbf{B}$
$$
\begin{aligned}
& \mathbf{E}=\mathbf{v} \times \mathbf{B} \\
& \mathbf{E} \times \mathbf{B}=(\mathbf{V} \times \mathbf{B}) \times \mathbf{B} \\
& v \times \mathbf{B}=v B \hat{\mathbf{n}} \\
& \Rightarrow \quad(\mathbf{v} B) \times \mathbf{B}=\mathbf{v}(\mathbf{B} \cdot \mathbf{B}) \\
& \Rightarrow \quad \mathbf{v}=\frac{\mathbf{E} \times \mathbf{B}}{\mathbf{B} \cdot \mathbf{B}} \\
& \mathbf{v} \times \mathbf{E}=\mathbf{v} \times \mathbf{v} \times \mathbf{B}=0 \neq \mathbf{B} \\
&
\end{aligned}
$$
[iii) is right]
Now, $\quad \mathbf{E}=\mathbf{v} \times \mathbf{B}$
$$
\mathbf{v} \times \mathbf{E}=\mathbf{v} \times \mathbf{v} \times \mathbf{B}=0 \neq \mathbf{B}
$$
So, option (d) is wrong.
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