Magnetics Ques 124
- A uniform magnetic field with a slit system as shown in figure is to be used as momentum filter for high-energy charged particles. With a field $B$ Tesla, it is found that the filter transmits $\alpha$-particles each of energy 5.3 MeV. The magnetic field is increased to $2.3 B$ Tesla and deuterons are passed into the filter. The energy of each deuteron transmitted by the filter is …… MeV.
(1997C, 1M)
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Answer:
Correct Answer: 124.14.0185
Solution:
Formula:
Magnetic Force Acting On A Moving Point Charge:
- Radius of circular path is given by
$$ R=\frac{m v}{B q}=\frac{p}{B q}=\frac{\sqrt{2 K m}}{B q} $$
Radius in both the cases are equal. Therefore,
$$ \begin{aligned} \frac{\sqrt{2 K _{\alpha} m _{\alpha}}}{B q _{\alpha}} & =\frac{\sqrt{2 K _d m _d}}{2.3 B q _d}, \\ \frac{q _d}{q _{\alpha}} & =\frac{e}{2 e}=\frac{1}{2} \\ \frac{m _{\alpha}}{m _d} & =\frac{4}{2}=2 \\ \therefore \quad K _d & =\frac{2.3 q _d}{q _{\alpha}} \frac{m _{\alpha}}{m _d} \cdot K _{\alpha} \\ K _d & =2.3 \times \frac{1}{2}(2)(5.3) MeV, \\ K _d & =14.0185 MeV \end{aligned} $$
BETA
