A charge moving with velocity `V` in `X` direction is subjected to a field of magnetic induction in the negative `X` direction . As a result the charge will

Options

A. remain unaffected
B. start moving in a circular y-z plane
C. retard along x-axis
D. move along a helical path around x-axis

Answer

To solve the problem, we will analyze the situation step by step using the concepts of magnetic force on a moving charge.

### Step 1: Understand the scenario
A charge \( q \) is moving with a velocity \( \mathbf{V} \) in the positive \( x \)-direction. The magnetic field \( \mathbf{B} \) is directed in the negative \( x \)-direction.

### Step 2: Identify the vectors
- The velocity vector \( \mathbf{V} \) can be represented as:
\[ \mathbf{V} = V \hat{i} \]
where \( \hat{i} \) is the unit vector in the positive \( x \)-direction.

- The magnetic field vector \( \mathbf{B} \) can be represented as:
\[ \mathbf{B} = -B \hat{i} \]
where \( B \) is the magnitude of the magnetic field and \( -\hat{i} \) indicates the negative \( x \)-direction.

### Step 3: Apply the formula for magnetic force
The force \( \mathbf{F} \) on a charge moving in a magnetic field is given by the equation:
\[ \mathbf{F} = q (\mathbf{V} \times \mathbf{B}) \]

### Step 4: Calculate the cross product
To find the force, we need to compute the cross product \( \mathbf{V} \times \mathbf{B} \):
\[ \mathbf{V} \times \mathbf{B} = (V \hat{i}) \times (-B \hat{i}) \]

### Step 5: Evaluate the cross product
The cross product of any vector with itself is zero:
\[ \hat{i} \times \hat{i} = 0 \]
Thus,
\[ \mathbf{V} \times \mathbf{B} = V \cdot (-B) \cdot (\hat{i} \times \hat{i}) = 0 \]

### Step 6: Determine the force
Substituting the result of the cross product back into the force equation:
\[ \mathbf{F} = q \cdot 0 = 0 \]

### Step 7: Conclusion
Since the force acting on the charge is zero, the charge will continue to move unaffected in the positive \( x \)-direction.

### Final Answer
The charge will move unaffected.

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