An electron of mass `0.90 xx 10^(-30)` kg under the action of a magnetic field moves in a circle of 2.0 cm radius at a speed `3.0 xx 10^6 ms^-1`. If a proton of mass `1.8 xx 10^(-27) kg` was to move in a circle of the same radius in the same magnetic field, then its speed will be

Options

A. `1.5 xx 10^(3) m//s`
B. `3 xx 10^(6) m//s`
C. `6 xx 10^(4) m//s`
D. `2 xx 10^(8) m//s`

Answer

To solve the problem, we will use the relationship between the mass, speed, charge, magnetic field, and radius of the circular motion of charged particles in a magnetic field. The formula we will use is:

\[ mv = qBr \]

Where:
- \( m \) = mass of the particle
- \( v \) = speed of the particle
- \( q \) = charge of the particle
- \( B \) = magnetic field strength
- \( r \) = radius of the circular path

### Step-by-Step Solution:

1. **Identify the known values for the electron:**
- Mass of electron, \( m_e = 0.90 \times 10^{-30} \) kg
- Speed of electron, \( v_e = 3.0 \times 10^6 \) m/s
- Radius of circular motion, \( r = 2.0 \) cm = \( 0.02 \) m
- Charge of electron, \( q_e = -1.6 \times 10^{-19} \) C (magnitude only)

2. **Identify the known values for the proton:**
- Mass of proton, \( m_p = 1.8 \times 10^{-27} \) kg
- Charge of proton, \( q_p = +1.6 \times 10^{-19} \) C (magnitude only)

3. **Since the radius \( r \) and the magnetic field \( B \) are the same for both particles, we can set up the equation:**
\[ m_e v_e = m_p v_p \]

4. **Rearranging the equation to find the speed of the proton \( v_p \):**
\[ v_p = \frac{m_e v_e}{m_p} \]

5. **Substituting the known values into the equation:**
\[ v_p = \frac{(0.90 \times 10^{-30} \text{ kg}) \times (3.0 \times 10^6 \text{ m/s})}{1.8 \times 10^{-27} \text{ kg}} \]

6. **Calculating the numerator:**
\[ 0.90 \times 10^{-30} \times 3.0 \times 10^6 = 2.7 \times 10^{-24} \text{ kg m/s} \]

7. **Calculating the speed of the proton:**
\[ v_p = \frac{2.7 \times 10^{-24}}{1.8 \times 10^{-27}} = 1.5 \times 10^3 \text{ m/s} \]

### Final Answer:
The speed of the proton \( v_p \) is \( 1.5 \times 10^3 \) m/s.

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