A proton with 1MeV kinetic energy is moving in a circular path of radius R in a uniform magnetic field. What should be the energy of an α – particle to describe a circle of same radius in the same magnetic field?a)2 MeVb)0.5 MeVc)1 MeVd)4 MeVCorrect answer is option 'C'. Can you explain this answer?

Question

A proton with 1MeV kinetic energy is moving in a circular path of radius R in a uniform magnetic field. What should be the energy of an &#945; &#8211; particle to describe a circle of same radius in the same magnetic field?&#8203;a)2 MeVb)0.5 MeVc)1 MeVd)4 MeVCorrect answer is option 'C'. Can you explain this answer?

Answer

Answer

According to the principal of circular motion in a magnetic field

Fc = Fm ⇒ mv^2/R = qvB

⇒ R = mv/qB = P/qB = √2m.k/qB

Rα = √2(4m)K’/2qB

R/Rα =√ K/K’

but R =Rα (given) Thus K=K’=1 MeV

Answer

Radius of proton= radius of alpha
mpVP/BQp=malpha.Valpha/BQalpha
Magnetic field is the same for both
Qalpha= 2Qp. malpha=4mp
mpVp=(4mp.Valpha)QP/2Qp
Valpha=VP/2
K.Ep=1Mev
K.Ealpha=1/2malpha(Valpha)^2
=1/2(4mp)(Vp/2)^2
=1/2mpvp^2=1Mev

Answer

Explanation:

When a charged particle moves in a magnetic field, it experiences a force called the magnetic force. The magnetic force can change the direction of the particle's motion but does not change its speed. In a uniform magnetic field, the magnetic force always acts perpendicular to the velocity of the particle, causing it to move in a circular path.

The magnetic force acting on a charged particle moving in a magnetic field can be calculated using the equation:

F = qvB sin(θ)

Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field strength
θ is the angle between the velocity and the magnetic field

In this case, the proton has a kinetic energy of 1 MeV, which can be converted to its velocity using the equation:

KE = 1/2 mv^2

Where:
KE is the kinetic energy
m is the mass of the proton
v is the velocity of the proton

Since the mass of a proton is approximately 1.67 x 10^-27 kg, we can solve for v:

1 MeV = 1/2 (1.67 x 10^-27 kg) v^2
v^2 = (2 x 1 MeV) / (1.67 x 10^-27 kg)
v^2 ≈ 1.20 x 10^19 m^2/s^2
v ≈ √(1.20 x 10^19) m/s
v ≈ 1.10 x 10^9 m/s

The proton is moving in a circular path of radius R. The centripetal force required to keep the proton moving in this circular path is provided by the magnetic force:

F = mv^2 / R = qvB sin(θ)

Since the velocity of the proton is perpendicular to the magnetic field, the angle θ is 90 degrees and sin(θ) = 1:

mv^2 / R = qvB

The mass of the proton cancels out:

v^2 / R = qB

Solving for v:

v = √(qBR)

Now, we need to find the velocity of the particle that would produce a circular path of the same radius R in the same magnetic field. Let's call this velocity v':

v' = √(qBR)

To find the kinetic energy of the particle, we can use the equation:

KE = 1/2 mv'^2

Substituting the value of v':

KE = 1/2 m(qBR)^2

Since the mass and the charge of the particle are the same as the proton, the kinetic energy of the particle would be:

KE = 1/2 (1.67 x 10^-27 kg) (1.6 x 10^-19 C) (B) (R)^2

Simplifying this expression, we get:

KE = 1.07 x 10^-11 B R^2

Since the kinetic energy of the proton is 1 MeV, which is equivalent to 1.6 x 10^-13 J, we can set up the following equation:

1.6 x 10^-13 J = 1.07 x 10^-11 B R^2

Solving for B:

B = (1.6 x 10

Answer

Kinetic energy of a charged particle,

Radius of the circular path of a charged particle uniform magnetic field is given by

Mass of a proton, m_p = m

Mass of an α-particle. m_α = 4m

Charge of a proton, q_p = e

Charge of an α -particle, q_α = 2_e

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