Anonymous
Sep 4
A proton accelerated through a potential difference of 3.0kV enters a magnetic field of flux density 0.1, in a plane perpendicular to the motion of the proton. Calculate, radius of circular that the?
particle describes in the magnetic field?
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ChatGPT
Sep 4
To calculate the radius of the circular path that the proton will follow in the magnetic field, we can use the formula for the radius of a charged particle moving in a magnetic field:
r = (m*v) / (q*B)
Where:
r = radius of the circular path
m = mass of the proton = 1.67 x 10^-27 kg
v = velocity of the proton
q = charge of the proton = 1.6 x 10^-19 C
B = magnetic flux density = 0.1 T
First, we need to find the velocity of the proton. We can use the formula for the kinetic energy of a charged particle accelerated through a potential difference:
KE = q*V
Where:
KE = kinetic energy
q = charge of the proton
V = potential difference = 3.0 kV = 3.0 x 10^3 V
Since the kinetic energy gained by the proton is equal to the work done on it by the potential difference, we can equate the two:
1/2*m*v^2 = q*V
Solving for v:
v = sqrt((2*q*V) / m)
v = sqrt((2*1.6 x 10^-19 C * 3.0 x 10^3 V) / 1.67 x 10^-27 kg)
v = 1.73 x 10^7 m/s
Now we can plug in the values for v, m, q, and B into the formula for the radius of the circular path:
r = (m*v) / (q*B)
r = (1.67 x 10^-27 kg * 1.73 x 10^7 m/s) / (1.6 x 10^-19 C * 0.1 T)
r = 2.86 x 10^-2 m
Therefore, the radius of the circular path that the proton will follow in the magnetic field is 2.86 x 10^-2 meters.
r = (m*v) / (q*B)
Where:
r = radius of the circular path
m = mass of the proton = 1.67 x 10^-27 kg
v = velocity of the proton
q = charge of the proton = 1.6 x 10^-19 C
B = magnetic flux density = 0.1 T
First, we need to find the velocity of the proton. We can use the formula for the kinetic energy of a charged particle accelerated through a potential difference:
KE = q*V
Where:
KE = kinetic energy
q = charge of the proton
V = potential difference = 3.0 kV = 3.0 x 10^3 V
Since the kinetic energy gained by the proton is equal to the work done on it by the potential difference, we can equate the two:
1/2*m*v^2 = q*V
Solving for v:
v = sqrt((2*q*V) / m)
v = sqrt((2*1.6 x 10^-19 C * 3.0 x 10^3 V) / 1.67 x 10^-27 kg)
v = 1.73 x 10^7 m/s
Now we can plug in the values for v, m, q, and B into the formula for the radius of the circular path:
r = (m*v) / (q*B)
r = (1.67 x 10^-27 kg * 1.73 x 10^7 m/s) / (1.6 x 10^-19 C * 0.1 T)
r = 2.86 x 10^-2 m
Therefore, the radius of the circular path that the proton will follow in the magnetic field is 2.86 x 10^-2 meters.
0
Google Gemini AI
Sep 4
Given:
- Potential difference: V = 3.0 kV = 3000 V
- Magnetic flux density: B = 0.1 T
Kinetic Energy of the proton:
The kinetic energy (KE) gained by the proton from the potential difference is:
```
KE = qV
```
where q is the charge of the proton.
```
KE = (1.602 x 10^-19 C)(3000 V)
KE = 4.806 x 10^-16 J
```
Radius of the Circular Path:
The magnetic force on the proton provides the centripetal force for the circular motion. Therefore,
```
qvB = mv^2/r
```
where m is the mass of the proton, v is the velocity of the proton, and r is the radius of the circular path.
Solving for r:
Rearranging the equation and substituting KE for mv^2/2, we get:
```
r = (2KE) / (qvB)
```
```
r = (2 - 4.806 x 10^-16 J) / ((1.602 x 10^-19 C)(0.1 T))
r = 0.601 m
```
Therefore, the radius of the circular path traveled by the proton is approximately 0.601 m.
- Potential difference: V = 3.0 kV = 3000 V
- Magnetic flux density: B = 0.1 T
Kinetic Energy of the proton:
The kinetic energy (KE) gained by the proton from the potential difference is:
```
KE = qV
```
where q is the charge of the proton.
```
KE = (1.602 x 10^-19 C)(3000 V)
KE = 4.806 x 10^-16 J
```
Radius of the Circular Path:
The magnetic force on the proton provides the centripetal force for the circular motion. Therefore,
```
qvB = mv^2/r
```
where m is the mass of the proton, v is the velocity of the proton, and r is the radius of the circular path.
Solving for r:
Rearranging the equation and substituting KE for mv^2/2, we get:
```
r = (2KE) / (qvB)
```
```
r = (2 - 4.806 x 10^-16 J) / ((1.602 x 10^-19 C)(0.1 T))
r = 0.601 m
```
Therefore, the radius of the circular path traveled by the proton is approximately 0.601 m.
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