Question

A proton, a deuteron and an \( \alpha \) particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is \( \qquad \) and their speed is \( \qquad \) in the ratio. (1) \( 1: 2: 4 \) and \( 2: 1: 1 \) (2) \( 2: 1: 1 \) and \( 4: 2: 1 \) (3) \( 4: 2: 1 \) and \( 2: 1: 1 \) (4) \( 1: 2: 4 \) and \( 1: 1: 2 \)

Answer 1

Given Data,

The momentum in a Uniform Magnetic Field is the same for a proton, a deuteron, and α αparticle.

Let the momentum be p.

Let, F₁ is Magnetic Force of proton, F₂ is Magnetic Force of deuteron, F₃ is Magnetic Force of α particle.

The force is given by

F = qvB (where, F is Magnetic Force, q is charge, v is velocity, and B is Magnetic Field.)

Since, velocity = Momentum/mass ⇒ v = p/m (where, m is mass, and p is momentum.)

If M is mass of proton, then the mass of a deutron and a αparticle can be written as 2M and 4M respectively.

Putting value of Velocity, \(F= \frac{qpB}{m}\)

Force of Proton \(F_1 = \frac {qpB}{M}\)

Force of Deuteron \(F_2= \frac {qpB}{2M}\)

Force of αparticle \(F_3= \frac {qpB}{3M}\)

Then the ratio of Magnetic Forces acting on a proton, a deuteron, and a α-particle is,

\(F_1 :F_2 :F_3 = 1 : \frac 12 : \frac 12\)

\(\Rightarrow F_1 :F_2:F_3 = 2: 1:1\)

Velocity of Proton v₁ = \(\frac pM\)

Velocity of Deuteron v₂ = \(\frac{p}{2M}\)

Velocity of αparticle v₃ = \(\frac{p}{4M}\)

Then the ratio of Velocity of a proton, a deuteron, and a α-particle is,

\(V_1 : V_2 :V_3 = 1 : \frac 12 : \frac 14\)

\(\Rightarrow V_1 : V_2 :V_3 = 4:2:1\)