Question

An electron enters a uniform magnetic field with a path perpendicular to the field lines and moves in a circular path of radius $R$.
A positron (same mass of an electron but opposite charge) enters the same magnetic field in a path perpendicular to the field with four times the kinetic energy.
Compared to the electrons path, the radius of the protons circular path is:

• A. The same radius.
• B. Twice as large.
• C. 1/2 as large.
• D. 4 times larger.

Answer 1

Answer: (B)

Twice as large.

Let the velocity of the electron be $v$ and that of positron be $v^{\prime }$.

Thus radius of circular path traced by electron $R=\frac{mv}{B|q|}$

Kinetic energy of positron is 4 times that of electron i.e $E_p=4E_e$

$\therefore$ $\frac{1}{2}m(v^{\prime }{)}^2=4\times \frac{1}{2}m{v}^2$ $⟹v^{\prime }=2v$

$\therefore$ Radius of circular path traced by positron $R^{\prime }=\frac{m{v}^{\prime }}{B|q|}=\frac{m\left(2v\right)}{B|q|}=2R$

Thus option B is correct.