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- ais`. 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)`

Explanation

To solve the problem, we need to find the magnitude of the magnetic field \( B \) acting on a charged particle moving in an electric field. The steps to solve the problem are as follows:

### Step 1: Understand the Forces Acting on the Charged Particle
The charged particle experiences two forces: the electric force \( F_E \) due to the electric field \( E \) and the magnetic force \( F_B \) due to the magnetic field \( B \). For the particle to continue moving along the x-axis without deflection, these two forces must be equal in magnitude and opposite in direction.

### Step 2: Write the Expression for Electric Force
The electric force \( F_E \) acting on the charged particle can be expressed as:
\[ F_E = Q \cdot E \]
where \( Q \) is the charge of the particle and \( E \) is the magnitude of the electric field.

### Step 3: Write the Expression for Magnetic Force
The magnetic force \( F_B \) acting on the charged particle can be expressed as:
\[ F_B = Q \cdot v \cdot B \]
where \( v \) is the velocity of the particle and \( B \) is the magnitude of the magnetic field.

### Step 4: Set the Forces Equal
Since the particle is undeflected, we can set the magnitudes of the electric force and magnetic force equal to each other:
\[ Q \cdot E = Q \cdot v \cdot B \]

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

### Step 6: Solve for the Magnetic Field \( B \)
Rearranging the equation gives us:
\[ B = \frac{E}{v} \]

### Step 7: Substitute the Given Values
Now we can substitute the given values into the equation:
- \( E = 10^4 \, \text{N/C} \)
- \( v = 10 \, \text{m/s} \)

Thus,
\[ B = \frac{10^4}{10} = 10^3 \, \text{T} \]

### Step 8: Conclusion
The magnitude of the magnetic field \( B \) is \( 10^3 \, \text{T} \).

### Final Answer
The magnitude of \( B \) is \( 10^3 \, \text{T} \).

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