A particle of charge `-16xx10^(-18)` coulomb moving with velocity `10 ms^(-1)` along the ` x-` axis , and an electric field of magnitude `10^(4)//(m)` is along the negative ` z-` axis. If the charged particle continues moving along the ` x`- axis , the magnitude of `B` is

Options

A. `10^(3) "Wb m"^(-2)`
B. `10^(5) "Wb m"^(-2)`
C. `10^(16) "Wb m"^(-2)`
D. `10^(-3) "Wb m"^(-2)`

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Identify the forces acting on the charged particle
The charged particle experiences two forces:
1. The electric force due to the electric field.
2. The magnetic force due to the magnetic field.

### Step 2: Write the expression for the electric force
The electric force (\( F_E \)) acting on the charged particle can be calculated using the formula:
\[ F_E = Q \cdot E \]
where:
- \( Q = -16 \times 10^{-18} \, \text{C} \) (charge of the particle)
- \( E = 10^4 \, \text{N/C} \) (magnitude of the electric field)

### Step 3: Write the expression for the magnetic force
The magnetic force (\( F_B \)) acting on the charged particle can be calculated using the formula:
\[ F_B = Q \cdot v \cdot B \]
where:
- \( v = 10 \, \text{m/s} \) (velocity of the particle)
- \( B \) is the magnetic field we need to find.

### Step 4: Set the electric force equal to the magnetic force
Since the particle continues moving along the x-axis and does not experience any displacement in the z-direction, the electric force must be balanced by the magnetic force:
\[ F_E = F_B \]
This leads to the equation:
\[ Q \cdot E = Q \cdot v \cdot B \]

### Step 5: Cancel the charge \( Q \) from both sides
Since \( Q \) is non-zero, we can divide both sides of the equation by \( Q \):
\[ E = v \cdot B \]

### Step 6: Solve for the magnetic field \( B \)
Rearranging the equation gives us:
\[ B = \frac{E}{v} \]
Substituting the known values:
\[ B = \frac{10^4}{10} = 10^3 \, \text{T} \]

### Conclusion
The magnitude of the magnetic field \( B \) is \( 10^3 \, \text{T} \).