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The force $F$ acting on charge $q$ moving with velocity $v$ perpendicular to magnetic field $B$ is i) $\quad F=q v B$ ii) $\quad F=\frac{q v}{B}$ iii) $\quad F=\frac{q B}{v}$ iv) $\quad F=\frac{B v}{q}$
The force $F$ acting on charge $q$ moving with velocity $v$ perpendicular to magnetic field $B$ is
i) $\quad F=q v B$
ii) $\quad F=\frac{q v}{B}$
iii) $\quad F=\frac{q B}{v}$
iv) $\quad F=\frac{B v}{q}$
Physics | Previous Year Question Papers
The force $F$ acting on charge $q$ moving with velocity $v$ perpendicular to magnetic field $B$ is
i) $\quad F=q v B$
ii) $\quad F=\frac{q v}{B}$
iii) $\quad F=\frac{q B}{v}$
iv) $\quad F=\frac{B v}{q}$

Correct option is (I) $F=q v B$
Magnetic force $\mathrm{F}=\mathrm{q}(\mathrm{v} \times \mathrm{B})=\mathrm{qvB} \sin \theta$ where $\theta$ is the angle between $v$ and $B$.
Given: $\theta=90^{\circ}$
Thus, force $\mathrm{F}=\mathrm{qvB} \times \sin 90^{\circ}=\mathrm{qvB}$

i

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