> #mathematics , #physics > Calculate the radius of the path traversed by an electron of energy 450ev moving at a right angle to a uniform magnetic field of flux density 1.5×10^-3T?
###### Anonymous
more than 1 week ago

# Calculate the radius of the path traversed by an electron of energy 450ev moving at a right angle to a uniform magnetic field of flux density 1.5×10^-3T?

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ChatGPT
more than 1 week ago
The radius of the path traversed by an electron moving at a right angle to a uniform magnetic field can be calculated using the formula:

r = (m*v)/(q*B)

Where:
r = radius of the path
m = mass of the electron = 9.11 x 10^-31 kg
v = velocity of the electron
q = charge of the electron = 1.6 x 10^-19 C
B = magnetic field flux density = 1.5 x 10^-3 T

First, we need to calculate the velocity of the electron using the energy given:

Energy of the electron = 450 eV = 450 x 1.6 x 10^-19 J
Kinetic energy of the electron = 1/2 * m * v^2
450 x 1.6 x 10^-19 = 1/2 * 9.11 x 10^-31 * v^2
v^2 = (2 * 450 x 1.6 x 10^-19) / 9.11 x 10^-31
v = sqrt((2 * 450 x 1.6 x 10^-19) / 9.11 x 10^-31)

Now, we can substitute the values into the formula for the radius:

r = (9.11 x 10^-31 * v) / (1.6 x 10^-19 * 1.5 x 10^-3)
r = (9.11 x 10^-31 * sqrt((2 * 450 x 1.6 x 10^-19) / 9.11 x 10^-31)) / (1.6 x 10^-19 * 1.5 x 10^-3)

Calculating the value of r will give us the radius of the path traversed by the electron.
0
more than 1 week ago
Given:

- Electron energy: E = 450 eV = 450 × 1.6 × 10^-19 J
- Magnetic flux density: B = 1.5 × 10^-3 T

Formula:

The radius of the path traversed by an electron in a magnetic field is given by:

```
r = (mv) / (qB)
```

where:

- r is the radius of the path
- m is the electron mass (9.11 × 10^-31 kg)
- v is the electron speed
- q is the electron charge (1.6 × 10^-19 C)
- B is the magnetic flux density

Solution:

First, we need to find the electron's speed. We can use the equation:

```
E = (1/2)mv^2
```

Solving for v, we get:

```
v = √(2E / m)
```

Plugging in the given values for E and m, we get:

```
v = √(2 × 450 × 1.6 × 10^-19 J / 9.11 × 10^-31 kg) ≈ 9.19 × 10^5 m/s
```

Now we can calculate the radius of the path:

```
r = (mv) / (qB) = (9.11 × 10^-31 kg × 9.19 × 10^5 m/s) / (1.6 × 10^-19 C × 1.5 × 10^-3 T) ≈ 0.336 m
```

Therefore, the radius of the path traversed by the electron is approximately 0.336 m.
0
Collins Benton
more than 1 week ago
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