An electron having momentum `2.4 xx 10^(-23)" kg m s"^(-1)` enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of `30^(@)` with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be

Options

A. 2mm
B. 1mm
C. `sqrt3/2` mm
D. 0.5mm

Explanation

To find the radius of the helical path of an electron moving in a magnetic field, we can follow these steps:

### Step 1: Understand the motion of the electron
The electron enters a magnetic field at an angle of \(30^\circ\) to the field. The motion of the electron will be helical due to the combination of two components of its velocity: one component is perpendicular to the magnetic field (circular motion), and the other is parallel to the magnetic field (linear motion).

### Step 2: Resolve the momentum into components
Given the momentum \(p\) of the electron is \(2.4 \times 10^{-23} \, \text{kg m/s}\), we can resolve this momentum into two components:
- The perpendicular component \(p_{\perp} = p \sin \theta\)
- The parallel component \(p_{\parallel} = p \cos \theta\)

Where \(\theta = 30^\circ\).

### Step 3: Calculate the perpendicular momentum
Using the sine function:
\[ p_{\perp} = p \sin(30^\circ) = 2.4 \times 10^{-23} \times \frac{1}{2} = 1.2 \times 10^{-23} \, \text{kg m/s} \]

### Step 4: Use the formula for the radius of the circular motion
The radius \(r\) of the circular motion (which is part of the helical path) can be calculated using the formula:
\[ r = \frac{m v_{\perp}}{q B} \]
Where:
- \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)),
- \(v_{\perp}\) is the speed corresponding to the perpendicular momentum,
- \(q\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \(B\) is the magnetic field strength (\(0.15 \, \text{T}\)).

### Step 5: Relate momentum to velocity
The momentum \(p\) is related to velocity \(v\) by:
\[ p = mv \implies v = \frac{p}{m} \]
Thus, the perpendicular velocity can be calculated as:
\[ v_{\perp} = \frac{p_{\perp}}{m} = \frac{1.2 \times 10^{-23}}{9.11 \times 10^{-31}} \approx 1.32 \times 10^{7} \, \text{m/s} \]

### Step 6: Substitute values into the radius formula
Now substituting the values into the radius formula:
\[ r = \frac{(9.11 \times 10^{-31}) \times (1.32 \times 10^{7})}{(1.6 \times 10^{-19}) \times (0.15)} \]

### Step 7: Calculate the radius
Calculating the numerator and denominator:
- Numerator: \(9.11 \times 10^{-31} \times 1.32 \times 10^{7} \approx 1.20 \times 10^{-23}\)
- Denominator: \(1.6 \times 10^{-19} \times 0.15 \approx 2.4 \times 10^{-20}\)

Thus,
\[ r \approx \frac{1.20 \times 10^{-23}}{2.4 \times 10^{-20}} = 0.5 \times 10^{-3} \, \text{m} = 0.5 \, \text{mm} \]

### Final Answer
The radius of the helical path of the electron is \(0.5 \, \text{mm}\).

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