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. `3.0 xx 10^6 ms^-1`
B. `1.5 xx 10^3 ms^-1`
C. `6.0 xx 10^4 ms^-1`
D. cannot be estimated from the given data

Explanation

To solve the problem, we will use the relationship between the mass of the particle, the radius of the circular path, the speed of the particle, and the magnetic field. The centripetal force acting on the charged particle moving in a magnetic field is provided by the magnetic force.

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Particles**:
The magnetic force acting on a charged particle moving in a magnetic field provides the centripetal force required for circular motion. The equation for this is:
\[ \frac{mv^2}{r} = qvB \]
where \( m \) is the mass of the particle, \( v \) is its speed, \( r \) is the radius of the circular path, \( q \) is the charge of the particle, and \( B \) is the magnetic field strength.

2. **Set Up the Equation for the Electron**:
For the electron, we can express the relationship as:
\[ v_e = \frac{q_e B r}{m_e} \]
where \( v_e \) is the speed of the electron, \( q_e \) is the charge of the electron, \( m_e \) is the mass of the electron, and \( r \) is the radius of the circular path.

3. **Set Up the Equation for the Proton**:
Similarly, for the proton, we have:
\[ v_p = \frac{q_p B r}{m_p} \]
where \( v_p \) is the speed of the proton, \( q_p \) is the charge of the proton, and \( m_p \) is the mass of the proton.

4. **Relate the Speeds of the Electron and Proton**:
Since both particles are in the same magnetic field and have the same radius, we can divide the two equations:
\[ \frac{v_e}{v_p} = \frac{q_e/m_e}{q_p/m_p} \]
Rearranging gives us:
\[ v_p = \frac{m_p}{m_e} v_e \cdot \frac{q_e}{q_p} \]

5. **Substitute Known Values**:
- The mass of the electron \( m_e = 0.90 \times 10^{-30} \) kg
- The mass of the proton \( m_p = 1.8 \times 10^{-27} \) kg
- The speed of the electron \( v_e = 3.0 \times 10^6 \) m/s
- The charge of the electron \( q_e = 1.6 \times 10^{-19} \) C
- The charge of the proton \( q_p = 1.6 \times 10^{-19} \) C

Since \( q_e = q_p \), the ratio \( \frac{q_e}{q_p} = 1 \).

6. **Calculate the Speed of the Proton**:
Substituting the values into the equation:
\[ v_p = \frac{1.8 \times 10^{-27}}{0.90 \times 10^{-30}} \cdot 3.0 \times 10^6 \]
\[ v_p = 2 \cdot 3.0 \times 10^6 = 6.0 \times 10^6 \text{ m/s} \]

### Final Answer:
The speed of the proton \( v_p \) is \( 6.0 \times 10^6 \) m/s.