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Problem: A helium nucleus (m= 6.65 x 10^-27 kg and q = 2e)is initially at rest. It is accelerated through a potential difference delta V and then enters a region with a constant magnetic field of strength B= 4T. It enter perpendicular to the field and undergoes circular motion with a radius of 0.001 m. Determine delta V?
Use the formula for the Larmor radius to get the velocity, v. You will find it at
http://en.wikipedia.org/wiki/Gyroradius
Then compute the kinetic energy of the particle, KE = (1/2) m v^2
Use that to get deltaV
KE = q*(deltaV)
Show your work for further assistance
To determine the potential difference (ΔV), we can use the equations for the circular motion of a charged particle in a magnetic field.
The centripetal force acting on the helium nucleus is provided by the magnetic force, which can be written as:
F = qvB
where:
F = centripetal force (mv²/r)
q = charge of the helium nucleus (2e)
v = velocity of the helium nucleus inside the magnetic field
B = strength of the magnetic field (4 T)
m = mass of the helium nucleus (6.65 x 10^-27 kg)
r = radius of the circular path (0.001 m)
Now, we can equate the centripetal force to the magnetic force:
mv²/r = qvB
Simplifying the equation, we can solve for the velocity (v):
v = (qBr) / m
Substituting the given values, we have:
v = (2e * 4 T * 0.001 m) / (6.65 x 10^-27 kg)
Now, we can calculate ΔV using the kinetic energy formula:
ΔV = qΔV = 0.5mv²
Substituting the values, we get:
(qΔV) = 0.5m [(qB
To determine the potential difference (delta V), we can use the principles of circular motion and the relationship between the magnetic force and the centripetal force acting on a charged particle.
Here's how you can solve the problem step by step:
Step 1: Understand the concepts and equations involved.
- The force experienced by a charged particle moving through a magnetic field is given by F = qVB, where q is the particle's charge, V is its velocity, and B is the magnetic field strength.
- The centripetal force acting on a particle in circular motion is given by F = (mv^2)/r, where m is the particle's mass, v is its velocity, and r is the radius of the circular path.
Step 2: Determine the magnitude of the force acting on the helium nucleus.
Since the helium nucleus has a charge of q = 2e, where e is the elementary charge (1.602 x 10^-19 C), the force experienced by the helium nucleus in the magnetic field is:
F = (2e)(v)(B), where e = 1.602 x 10^-19 C and B = 4 T (tesla).
Step 3: Relate the force to the centripetal force.
The force acting on the helium nucleus in the magnetic field is responsible for providing the necessary centripetal force for circular motion. Therefore, we can equate the two forces:
(2e)(v)(B) = (mv^2)/r
Step 4: Solve for the velocity (v).
Rearrange the equation to solve for v:
v = r(Bq)/(2m)
Step 5: Substitute the known values into the equation.
Plug in the values given:
- Radius (r) = 0.001 m
- Magnetic field strength (B) = 4 T
- Charge (q) = 2e, where e = 1.602 x 10^-19 C
- Mass (m) = 6.65 x 10^-27 kg
v = (0.001 m)(4 T)(2e) / (2)(6.65 x 10^-27 kg)
Step 6: Calculate the velocity (v).
Multiply and divide to simplify the equation:
v = (0.001 m)(4 T)(2)(1.602 x 10^-19 C) / (2)(6.65 x 10^-27 kg)
v ≈ 1.205 x 10^7 m/s
Step 7: Calculate the potential difference (delta V).
Since the helium nucleus starts from rest, the initial kinetic energy is zero. When the helium nucleus is accelerated through the potential difference delta V, it gains kinetic energy. So we can equate the kinetic energy gain to the potential difference:
(1/2)mv^2 = q(delta V), where q is the charge (2e) and v is the velocity (1.205 x 10^7 m/s).
Step 8: Solve for the potential difference (delta V).
Rearrange the equation to solve for delta V:
delta V = (1/2m)(q)(v^2)
Step 9: Substitute the known values into the equation.
Plug in the values:
- Charge (q) = 2e, where e = 1.602 x 10^-19 C
- Mass (m) = 6.65 x 10^-27 kg
- Velocity (v) = 1.205 x 10^7 m/s
delta V = (1/2)(6.65 x 10^-27 kg)(2e)(1.205 x 10^7 m/s)^2
Step 10: Calculate the potential difference (delta V).
Multiply and raise to the power:
delta V = (1/2)(6.65 x 10^-27 kg)(2)(1.602 x 10^-19 C)(1.205 x 10^7 m/s)^2
delta V ≈ 1.43 x 10^5 V
Therefore, the potential difference (delta V) is approximately 1.43 x 10^5 volts.