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Two particles of masses $m_{1}=m, m_{2}=2 m$ and charges $q_{1}=q, q_{2}=2 q$ entered into uniform magnetic field. Find $F_{1} / F_{2}$ (force ratio).
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The correct answer is:
$\frac{1}{2}$
Force on a charge particle moving in a uniform magnetic field at angle $\theta$ is given by
$F=q v B \sin \theta \Rightarrow$ Hence, $q_{1}=q$ and $q_{2}=2 q$ $\therefore \quad \frac{F_{1}}{F_{2}}=\frac{q_{1}}{q_{2}}=\frac{q}{2 q}=\frac{1}{2}$
$\therefore \quad \frac{F_{1}}{F_{2}}=\frac{q_{1}}{q_{2}}=\frac{q}{2 q}=\frac{1}{2}$ (c) Given, $W=50 \mathrm{~J}$ and $q=25 \mathrm{C}$
Potential difference, $V=\frac{\text { Work done }(W)}{\text { Charge }(q)}$
$\Rightarrow \quad V=\frac{50}{25}=2 \mathrm{~V}$
$F=q v B \sin \theta \Rightarrow$ Hence, $q_{1}=q$ and $q_{2}=2 q$ $\therefore \quad \frac{F_{1}}{F_{2}}=\frac{q_{1}}{q_{2}}=\frac{q}{2 q}=\frac{1}{2}$
$\therefore \quad \frac{F_{1}}{F_{2}}=\frac{q_{1}}{q_{2}}=\frac{q}{2 q}=\frac{1}{2}$ (c) Given, $W=50 \mathrm{~J}$ and $q=25 \mathrm{C}$
Potential difference, $V=\frac{\text { Work done }(W)}{\text { Charge }(q)}$
$\Rightarrow \quad V=\frac{50}{25}=2 \mathrm{~V}$
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