}) $$

According to Newton's second law, $\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}$.

Therefore,

$$ \mathrm{m} \overrightarrow{\mathrm{a}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) $$

The resulting vector is always perpendicular to both $\vec{v}$ and $\vec{B}$.

Since $\overrightarrow{\mathrm{a}}$ is in the same direction as $\overrightarrow{\mathrm{F}}$, the acceleration vector $\overrightarrow{\mathrm{a}}$ must be perpendicular to the magnetic field vector $\overrightarrow{\mathrm{B}}$.

$$ \overrightarrow{\mathrm{a}} \perp \overrightarrow{\mathrm{~B}} $$

If two vectors are perpendicular, their dot product must be zero:

$$ \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~B}}=0 $$

The magnetic field is given as $\vec{B}=3 \hat{\imath}+2 \hat{\jmath}$ and $\vec{a}=4 \hat{\imath}-\frac{x}{2} \hat{\jmath}$ :

$$ \left(4 \hat{\imath}-\frac{x}{2} \hat{\jmath}\right) \cdot(3 \hat{\imath}+2 \hat{\jmath})=0 $$

$$\Rightarrow $$ $(4 \times 3)+\left(-\frac{x}{2} \times 2\right)=0$

$$\Rightarrow $$ $12-x=0$

$$\Rightarrow $$ $$ x=12 $$

Therefore, the value of x is 12 .

19:T7ef,

चुंबकीय क्षेत्र $\vec{B}$ में वेग $\vec{v}$ से गतिमान आवेश $q$ वाले कण पर लगने वाला बल ( $\vec{F}$ ) लोरेंत्ज़ बल नियम द्वारा दिया जाता है:

$$ \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) $$

न्यूटन के दूसरे नियम के अनुसार, $\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}$ है।

इसलिए,

$$ \mathrm{m} \overrightarrow{\mathrm{a}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) $$

परिणामी सदिश हमेशा $\vec{v

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