A proton, a deuteron and an α 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.

CONCEPT:

• The momentum is defined as the product of mass and velocity and it is written as;

            p = mv

           Here p is momentum, m is the mass and v is the velocity.

• The Lorentz force is written as;

           F = q (v × B)    or, F = qvB sinθ 

CALCULATIONS:

Let " m " be the mass of the proton and "q" is the charge of the proton.

As we know the deuteron is have twice the mass of the proton.

and the α  particle is having four times the mass of the proton and twice the charge of the proton.

As we know that momentum is written as;

p = mv ⇒v = \(\frac{p}{m}\)

For proton we have;

 \(v_p = \frac{p}{m_p}\)  -----(1)

For deuteron we have;

\(v_D = \frac{p}{m_D}\)

⇒ \(v_D = \frac{p}{2 m_p}\)    ----(2)

and for \(α\) particle we have;

\(v_{α} = \frac{p}{m_{α}}\)

⇒ \(v_{α} = \frac{p}{4m_{p}}\)   ----(3)

Now, take the ratio of equations (1), (2), and (3) we have;

\(v_p: v_D: v_{α} = 1 : \frac{1}{2} : \frac{1}{4}\)

⇒ \(v_p: v_D: v_{α }\) = 4 ∶ 2 ∶ 1

Now, the magnetic force is written as;

F = qvB

and we know v = \(\frac{p}{m}\)

∴ F = \( \frac{qpB}{m}\)

Now, for proton we have;

\(F_{p} = \frac{q_p p B}{m_p}\)   ----(4)

For Deuteron we have;

\(F_{D} = \frac{q_D p B}{m_D}\)

⇒ \(F_{D} = \frac{q_p p B}{2m_p}\)   ----(5)

and for α particle we have;

\(F_{α} = \frac{q_{α} p B}{m_{α}}\)

⇒ \(F_{α} = \frac{2q_{p} p B}{4m_{p}}\)

⇒ \(F_{α} = \frac{q_{p} p B}{2m_{p}}\)    ----(6)

The ratio of equations (4), (5), and (6) we have;

\(F_p : F_D :F_{α} = \frac{q_{p} p B}{m_{p}}: \frac{q_{p} p B}{2m_{p}}: \frac{q_{p} p B}{2m_{p}}\)

⇒ \(F_p : F_D :F_{α} = 1 : \frac{1}{2} : \frac{1}{2}\)

⇒ \(F_p : F_D :F_{α} \) = 2 ∶ 1 ∶ 1

Hence, option 1) is the correct answer.