An electron beam passes through a magnetic field of `2xx10^-3 weber//m^2` and an electric field of `1*0xx10^4Vm^-1` both acting simultaneously. The path of electrons remaining undeviated, calculate the speed of the electrons. If the electric field is removed what will be the radius of the electron path?

Explanation

To solve the problem step by step, we will first calculate the speed of the electrons when both the electric and magnetic fields are present, and then we will find the radius of the electron's path when the electric field is removed.

### Step 1: Understanding the Forces Acting on the Electron
When an electron moves through both an electric field (E) and a magnetic field (B) simultaneously, the forces due to these fields must balance each other for the electron's path to remain undeviated. The force due to the electric field is given by:
\[ F_E = qE \]
where \( q \) is the charge of the electron and \( E \) is the electric field strength.

The magnetic force acting on the electron is given by:
\[ F_B = qvB \]
where \( v \) is the speed of the electron and \( B \) is the magnetic field strength.

For the electron's path to remain undeviated, these two forces must be equal:
\[ qE = qvB \]

### Step 2: Canceling the Charge and Rearranging the Equation
Since the charge \( q \) is the same on both sides of the equation, we can cancel it out:
\[ E = vB \]

Now, we can rearrange this equation to solve for the speed \( v \):
\[ v = \frac{E}{B} \]

### Step 3: Substituting the Given Values
We are given:
- Electric field \( E = 1 \times 10^4 \, \text{Vm}^{-1} \)
- Magnetic field \( B = 2 \times 10^{-3} \, \text{Wb/m}^2 \)

Substituting these values into the equation:
\[ v = \frac{1 \times 10^4}{2 \times 10^{-3}} \]

### Step 4: Calculating the Speed
Now, we perform the calculation:
\[ v = \frac{1 \times 10^4}{2 \times 10^{-3}} = \frac{1 \times 10^4}{0.002} = 5 \times 10^6 \, \text{m/s} \]

### Step 5: Finding the Radius of the Electron's Path When the Electric Field is Removed
When the electric field is removed, the electron will move in a circular path due to the magnetic field. The radius \( r \) of the circular path can be calculated using the formula:
\[ r = \frac{mv}{qB} \]
where:
- \( m \) is the mass of the electron \( (9.1 \times 10^{-31} \, \text{kg}) \)
- \( q \) is the charge of the electron \( (1.6 \times 10^{-19} \, \text{C}) \)

### Step 6: Substituting Values to Find the Radius
Substituting the known values into the radius formula:
\[ r = \frac{(9.1 \times 10^{-31}) (5 \times 10^6)}{(1.6 \times 10^{-19})(2 \times 10^{-3})} \]

### Step 7: Performing the Calculation
Calculating the numerator:
\[ 9.1 \times 10^{-31} \times 5 \times 10^6 = 4.55 \times 10^{-24} \]

Calculating the denominator:
\[ 1.6 \times 10^{-19} \times 2 \times 10^{-3} = 3.2 \times 10^{-22} \]

Now, calculating the radius:
\[ r = \frac{4.55 \times 10^{-24}}{3.2 \times 10^{-22}} \approx 0.0142 \, \text{m} = 1.42 \, \text{cm} \]

### Final Answers
1. The speed of the electrons is \( 5 \times 10^6 \, \text{m/s} \).
2. The radius of the electron's path when the electric field is removed is approximately \( 1.42 \, \text{cm} \).

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