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A proton, a deuteron and an $\alpha$ particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is.......... and their speed is.................. in the ratio.
$1: 2: 4$ and $2: 1: 1$
$2: 1: 1$ and $4: 2: 1$
$4: 2: 1$ and $2: 1: 1$
$1: 2: 4$ and $1: 1: 2$
Solution
$F=q(\vec{v} \times \vec{B})=\frac{q}{m}(\vec{P} \times \vec{B})$
$\Rightarrow F \propto \frac{ q }{ m }$
thus $F _{1}: F _{2}: F _{3}=\frac{ q _{1}}{ m _{1}}: \frac{ q _{2}}{ m _{2}}: \frac{ q _{3}}{ m _{3}}$
$=\frac{ e }{ m _{ p }}: \frac{ e }{2 m _{ p }}: \frac{2 e }{4 m _{ p }}$
$=\frac{1}{1}: \frac{1}{2}: \frac{2}{4}$
$=2: 1: 1$
Now for speed calculation
$P = constant \Rightarrow v \propto \frac{1}{ m }$
thus $v _{1}: v _{2}: v _{3}=\frac{1}{ m _{ p }}: \frac{1}{2 m _{ p }}: \frac{1}{4 m _{ p }}$
$=\frac{1}{1}: \frac{1}{2}: \frac{1}{4}$
$=4: 2: 1$
