Question

Two charged particles of specific charges in the ratio 2:1 and masses in the ratio 1:4 moving with same kinetic energy enter a uniform magnetic field at right angles to the direction of the field. The ratio of the radii of the circular paths in which the particles move under the influence of the magnetic field is

• A. 2:1
• B. 1:1
• C. 4:1
• D. 8:1

Hint:

Radius in magnetic field: $r = \frac{mv}{qB}$. Specific charge: $\frac{q}{m}$. Equal kinetic energy: $\frac{1}{2}mv^2$ is constant.

Answer 1

The Correct Option is B

The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$, where $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength. Specific charge is the ratio $\frac{q}{m}$. Let the specific charges be $s_1 = \frac{q_1}{m_1}$ and $s_2 = \frac{q_2}{m_2}$, with $s_1:s_2 = 2:1$. Let the masses be $m_1$ and $m_2$, with $m_1:m_2 = 1:4$. Since the kinetic energies are equal, $\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2$, so $\frac{v_1^2}{v_2^2} = \frac{m_2}{m_1} = \frac{4}{1}$. Thus, $v_1 = 2v_2$. $r_1 = \frac{m_1v_1}{q_1B}$ and $r_2 = \frac{m_2v_2}{q_2B}$. $\frac{r_1}{r_2} = \frac{m_1v_1/q_1}{m_2v_2/q_2} = \frac{v_1/s_1}{v_2/s_2} = \frac{v_1}{v_2} \times \frac{s_2}{s_1} = \frac{2v_2}{v_2} \times \frac{1}{2} = 1$. Therefore, $r_1:r_2 = 1:1$.