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(B) The magnetic force is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since $\vec{F}$ is perpendicular to $\vec{v}$, their dot product must be zero

Vedclass · Accepted Answer

$\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$ with $\frac{v_1}{v_2} = \frac{B_1}{B_2}$

Vedclass · Answer

(B) The magnetic force is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since $\vec{F}$ is perpendicular to $\vec{v}$, their dot product must be zero: $\vec{F} \cdot \vec{v} = 0$.$(F_1 \hat{i} + F_2 \hat{j}) \cdot (v_1 \hat{i} + v_2 \hat{j}) = F_1 v_1 + F_2 v_2 = 0 \Rightarrow \frac{F_1}{F_2} = -\frac{v_2}{v_1} \quad (I)$.Also, $\vec{F}$ is perpendicular to $\vec{B}$, so $\vec{F} \cdot \vec{B} = 0$.Let $\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$. Then $(F_1 \hat{i} + F_2 \hat{j}) \cdot (B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}) = F_1 B_1 + F_2 B_2 = 0 \Rightarrow \frac{F_1}{F_2} = -\frac{B_2}{B_1} \quad (II)$.Comparing $(I)$ and $(II)$, we get $\frac{v_2}{v_1} = \frac{B_2}{B_1}$, which implies $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.Thus, $\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$ satisfies the condition $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.

$A$ charged particle moving with a velocity $\vec{v} = v_1 \hat{i} + v_2 \hat{j}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F} = F_1 \hat{i} + F_2 \hat{j}$. Here $v_1, v_2, F_1, F_2$ are all constants. Then $\vec{B}$ can be

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