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 solve the problem step by step, we need to find the mass of a charge moving in a magnetic field and then transitioning to a straight line motion due to an electric field.

### Step 1: Understand the Forces Acting on the Charge
When the charge is moving in a circle, the magnetic force acts as the centripetal force. The magnetic force \( F_B \) can be expressed as:
\[ F_B = QVB \]
where:
- \( Q \) is the charge,
- \( V \) is the velocity of the charge,
- \( B \) is the magnetic field strength.

### Step 2: Set Up the Centripetal Force Equation
The centripetal force \( F_C \) required to keep the charge moving in a circle is given by:
\[ F_C = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the charge,
- \( v \) is the velocity,
- \( r \) is the radius of the circular path.

### Step 3: Equate the Forces
Since the magnetic force provides the centripetal force, we can set them equal to each other:
\[ QVB = \frac{mv^2}{r} \]

### Step 4: Rearranging for Velocity
From the equation above, we can express the velocity \( V \):
\[ V = \frac{QBr}{m} \]

### Step 5: Introduce the Electric Field
When the electric field \( E \) is applied, the electric force \( F_E \) acting on the charge is given by:
\[ F_E = EQ \]
For the charge to start moving in a straight line, the electric force must balance the magnetic force:
\[ EQ = QVB \]
This simplifies to:
\[ E = VB \]

### Step 6: Substitute for Velocity
From the earlier expression for \( V \):
\[ E = \frac{QBr}{m}B \]
This can be rearranged to find the mass \( m \):
\[ m = \frac{Q B^2 r}{E} \]

### Step 7: Plug in the Values
Now, substituting the known values:
- \( Q = 1.6 \times 10^{-19} \, C \)
- \( B = 0.5 \, T \)
- \( r = 5 \, cm = 0.05 \, m \)
- \( E = 0.15 \, V/m \)

Substituting these values into the equation:
\[ m = \frac{(1.6 \times 10^{-19}) \times (0.5)^2 \times (0.05)}{0.15} \]

### Step 8: Calculate the Mass
Calculating the components:
\[ B^2 = (0.5)^2 = 0.25 \]
Now substituting:
\[ m = \frac{(1.6 \times 10^{-19}) \times 0.25 \times 0.05}{0.15} \]
Calculating the numerator:
\[ 1.6 \times 0.25 \times 0.05 = 0.02 \times 10^{-19} \]
Now dividing by \( 0.15 \):
\[ m = \frac{0.02 \times 10^{-19}}{0.15} = \frac{2 \times 10^{-21}}{1.5} = \frac{4}{3} \times 10^{-20} \]

### Final Answer
Thus, the mass of the charge is:
\[ m = \frac{4}{3} \times 10^{-20} \, kg \]

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