> #mechanics , #physics > A beam of protons moves in circle of radius 0.25m. the beam moves perpendicular to a 0.3 T magnetic field. a) What is the speed of the protons? b) Determine the magnitude of the centripetal force?
###### Anonymous
more than 3 weeks ago

# A beam of protons moves in circle of radius 0.25m. the beam moves perpendicular to a 0.3 T magnetic field. a) What is the speed of the protons? b) Determine the magnitude of the centripetal force?

that acts on the proton?
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ChatGPT
more than 3 weeks ago
a) To find the speed of the protons, we can use the formula for the centripetal force acting on a charged particle moving in a magnetic field:

F = qvB

Where:
F = centripetal force
q = charge of the particle (proton charge = 1.6 x 10^-19 C)
v = velocity of the particle
B = magnetic field strength (0.3 T)

The centripetal force is provided by the magnetic force acting on the protons:

F = mv^2 / r

Setting these two equations equal to each other and solving for v:

qvB = mv^2 / r
v = qBr / m

Plugging in the values:

v = (1.6 x 10^-19 C)(0.3 T)(0.25 m) / (1.67 x 10^-27 kg)
v = 7.2 x 10^6 m/s

Therefore, the speed of the protons is 7.2 x 10^6 m/s.

b) To determine the magnitude of the centripetal force, we can use the formula:

F = mv^2 / r

Plugging in the values:

F = (1.67 x 10^-27 kg)(7.2 x 10^6 m/s)^2 / 0.25 m
F = 3.02 x 10^-13 N

Therefore, the magnitude of the centripetal force acting on the protons is 3.02 x 10^-13 N.
0
more than 3 weeks ago
a) Speed of the protons:

The magnetic force on a single proton is given by:

```
F = qvB
```

where:

- F is the magnetic force
- q is the charge of the proton (1.6 × 10^-19 C)
- v is the speed of the proton
- B is the magnetic field strength (0.3 T)

The centripetal force required to keep the proton moving in a circle is given by:

```
F_c = mv^2/r
```

where:

- F_c is the centripetal force
- m is the mass of the proton (1.67 × 10^-27 kg)
- v is the speed of the proton
- r is the radius of the circle (0.25 m)

Equating the magnetic force and the centripetal force, we get:

```
qvB = mv^2/r
```

Solving for v, we get:

```
v = qBr/m
```

Substituting the given values, we get:

```
v = (1.6 × 10^-19 C) × (0.3 T) × (0.25 m) / (1.67 × 10^-27 kg) = 3.00 × 10^6 m/s
```

Therefore, the speed of the protons is 3.00 × 10^6 m/s.

b) Magnitude of the centripetal force:

From the equation for centripetal force, we have:

```
F_c = mv^2/r
```

Substituting the given values, we get:

```
F_c = (1.67 × 10^-27 kg) × (3.00 × 10^6 m/s)^2 / (0.25 m) = 1.50 × 10^-14 N
```

Therefore, the magnitude of the centripetal force is 1.50 × 10^-14 N.
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