An Alpha particle and a Proton moving with the same speed enter the sa...
force acting on a particle entering at right angle to the magnetic field is equal to: Bvq
and this is equal to the centripetal force of the particle due to its rotation = mv2/r
thus radius = r = mv/Bq
now easily the ratio can be derieved.
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An Alpha particle and a Proton moving with the same speed enter the sa...
Alpha Particle and Proton in a Magnetic Field
The trajectory followed by an Alpha particle and a Proton in a magnetic field can be determined by applying the Lorentz force equation, which states that the force experienced by a charged particle in a magnetic field is perpendicular to both the velocity of the particle and the magnetic field.
Trajectory of Alpha Particle and Proton
When an Alpha particle and a Proton enter the magnetic field region at right angles to the direction of the field, they will experience a magnetic force that causes them to move in a circular path. The radius of the circular path followed by a charged particle in a magnetic field is given by the equation:
\[ r = \frac{mv}{qB} \]
where r is the radius of the circular path, m is the mass of the particle, v is the speed of the particle, q is the charge of the particle, and B is the magnetic field strength.
Ratio of Radii of Circular Paths
Since the Alpha particle and the Proton have the same speed and enter the magnetic field region at right angles to the field, the ratio of the radii of the circular paths which the two particles describe can be calculated as:
\[ \frac{r_{\text{Alpha}}}{r_{\text{Proton}}} = \frac{m_{\text{Alpha}}}{m_{\text{Proton}}} \times \frac{q_{\text{Proton}}}{q_{\text{Alpha}}} \]
Given that the mass of an Alpha particle is 4 times the mass of a Proton and the charge of an Alpha particle is twice the charge of a Proton, the ratio of the radii of the circular paths will be:
\[ \frac{r_{\text{Alpha}}}{r_{\text{Proton}}} = \frac{4}{1} \times \frac{1}{2} = 2 \]
Therefore, the ratio of the radii of the circular paths followed by the Alpha particle and the Proton in the magnetic field is 2.
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