First slide

Magnetic field due to a moving charge

Question

A proton of energy 8eV is moving in a circular path in a uniform magnetic field. The energy of an $α -$ particle moving in the same magnetic field and along the same path will be-

Easy

A

4eV
B

2eV
C

8eV
D

6eV

Solution

$r = \frac{mv}{qB} = \frac{p}{qB} = \frac{\sqrt{2 mk}}{qB}$ $\frac{p^{2}}{2 m} = k; p = \sqrt{2 mk}$
It r and B are same
$k \propto \frac{q}{\sqrt{m}} \frac{k_{p}}{k_{α}} = \frac{q}{2 q} \sqrt{\frac{4 m}{m} = 1} k_{p} = k_{α}$ .

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