A chance of magnitude `1.6 xx 10 ^(-19)C` moving in a circle of radius 5 cm in a uniform magnetic field 0.5 T After applying `E = 0. 15 V m^(-1)` charge starts moving in a straight line mass of charge will be

Options

A. `2/3xx10^(-20)kg`
B. `4/3xx10^(-20)kg`
C. `6/3xx10^(-20)kg`
D. `1/3xx10^(-20)kg`

Answer

To find the mass of the charged particle, we can follow these steps:

### Step 1: Understand the relationship between electric and magnetic forces
When a charged particle moves in a magnetic field and starts moving in a straight line after an electric field is applied, the electric force (F_e) and magnetic force (F_m) acting on it must be equal.

The electric force is given by:
\[ F_e = qE \]
where \( q \) is the charge and \( E \) is the electric field.

The magnetic force is given by:
\[ F_m = qvB \]
where \( v \) is the velocity of the charge and \( B \) is the magnetic field.

### Step 2: Set the forces equal
Since \( F_e = F_m \):
\[ qE = qvB \]

### Step 3: Cancel \( q \) from both sides
Assuming \( q \neq 0 \), we can cancel \( q \):
\[ E = vB \]

### Step 4: Solve for velocity \( v \)
From the equation \( E = vB \), we can express the velocity \( v \):
\[ v = \frac{E}{B} \]

### Step 5: Use the radius of the circular motion
The radius \( r \) of the circular motion of the charge is given by:
\[ r = \frac{mv}{qB} \]

### Step 6: Substitute \( v \) into the radius equation
Substituting \( v \) from Step 4 into the radius equation:
\[ r = \frac{m \left(\frac{E}{B}\right)}{qB} \]
\[ r = \frac{mE}{qB^2} \]

### Step 7: Rearranging to find mass \( m \)
Rearranging the equation to solve for mass \( m \):
\[ m = \frac{qBr}{E} \]

### Step 8: Substitute the known values
Now, substitute the given values into the equation:
- Charge \( q = 1.6 \times 10^{-19} \, C \)
- Magnetic field \( B = 0.5 \, T \)
- Radius \( r = 5 \, cm = 5 \times 10^{-2} \, m \)
- Electric field \( E = 0.15 \, V/m \)

\[ m = \frac{(1.6 \times 10^{-19} \, C)(0.5 \, T)(5 \times 10^{-2} \, m)}{0.15 \, V/m} \]

### Step 9: Calculate the mass
Calculating the numerator:
\[ (1.6 \times 10^{-19})(0.5)(5 \times 10^{-2}) = 4 \times 10^{-21} \]

Now divide by \( 0.15 \):
\[ m = \frac{4 \times 10^{-21}}{0.15} = \frac{4 \times 10^{-21}}{\frac{15}{100}} = \frac{4 \times 10^{-21} \times 100}{15} = \frac{400 \times 10^{-21}}{15} \]

Calculating:
\[ m = \frac{400}{15} \times 10^{-21} \approx 26.67 \times 10^{-21} \, kg \]

### Step 10: Final result
Expressing in scientific notation:
\[ m \approx 2.67 \times 10^{-20} \, kg \]

### Conclusion
Thus, the mass of the charged particle is approximately:
\[ m \approx 2.67 \times 10^{-20} \, kg \]