A charged particle is observed traveling in a circular path of radius R in a uniform magnetic field. If the particle was traveling twice as fast, the radius of the circular path would be
8R.
R/4.
R/2.
2R.
4R.
expanation lso
r = mv/qB
double r ---> double v
To determine how the radius of the circular path changes when the speed of the charged particle is doubled, we can use the equation for the magnetic force on a charged particle moving in a magnetic field:
F = qvB
Where:
F is the magnetic force on the particle,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field strength.
The centripetal force that keeps the particle moving in a circular path is given by:
F = (mv²) / R
Where:
m is the mass of the particle,
v is the velocity of the particle,
R is the radius of the circular path.
Since the magnetic force is responsible for providing the centripetal force, we can equate the two equations:
qvB = (mv²) / R
Canceling out the charge and mass from both sides, we get:
vB = v² / R
Rearranging the equation, we have:
v² = RB
Now, if the speed of the particle is doubled, we have:
(2v)² = RB
4v² = RB
Dividing both sides by B, we get:
4v² / B = R
Since R is the radius of the circular path, we find that the radius would be 4R when the particle is traveling twice as fast. Therefore, the correct answer is 4R.
To determine the relationship between the speed of the particle and the radius of the circular path, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field:
F = q * v * B * sin(theta)
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
- theta is the angle between the velocity vector and the magnetic field vector.
Since the particle is moving in a circular path, its velocity vector is always perpendicular to the magnetic field vector. Therefore, sin(theta) = 1.
In a circular path, the magnetic force provides the necessary centripetal force to keep the particle in the circular motion. The magnitude of the centripetal force is given by:
F_c = m * (v^2 / r)
where:
- F_c is the centripetal force
- m is the mass of the particle
- v is the velocity of the particle
- r is the radius of the circular path.
Since the centripetal force is provided by the magnetic force, we can equate the two equations:
m * (v^2 / r) = q * v * B
Canceling out the common factors, we get:
m * (v / r) = q * B
From this equation, we can see that the velocity of the particle is inversely proportional to the radius of the circular path. So, if the velocity of the particle is doubled, the radius of the circular path will be halved.
Therefore, the correct answer is R/2.