A positive charge 'q' of mass 'm' is moving along the +x axis. We wish to apply a uniform magnetic field B for time `Delta `t so that the charge reverses its direction crossing the y axis at a distance d. Then :

Options

A. `b=(m upsilon )/(2 qd) and Delta t = ( pi d )/( 2 upsilon )`
B. `B=(m upsilon )/( qd) and Delta t = (pi d )/( upsilon)`
C. `B= (2m upsilon )/( qd) and Delta t = (pi d )/( upsilon )`
D. `B=( 2m upsilon )/(qd ) and Delta t = ( pi d)/( 2 upsilon )`

Answer

To solve the problem, we need to find the expressions for the magnetic field \( B \) and the time interval \( \Delta t \) required for a positive charge \( q \) of mass \( m \) moving along the positive x-axis to reverse its direction after crossing the y-axis at a distance \( d \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The charge \( q \) is initially moving in the positive x-direction.
- We want it to reverse direction after crossing the y-axis at a distance \( d \).

2. **Path of the Charge**:
- The charge will follow a semicircular path due to the magnetic field applied perpendicular to its motion.
- Let the radius of the semicircular path be \( r \). The distance \( d \) is equal to the diameter of the semicircle, so:
\[ d = 2r \quad \Rightarrow \quad r = \frac{d}{2} \]

3. **Equating Forces**:
- The magnetic force acting on the charge is given by:
\[ F_B = qvB \]
- The centripetal force required to keep the charge moving in a circular path is given by:
\[ F_C = \frac{mv^2}{r} \]
- For the charge to move in a circular path, these forces must be equal:
\[ qvB = \frac{mv^2}{r} \]

4. **Solving for Magnetic Field \( B \)**:
- Rearranging the above equation gives:
\[ B = \frac{mv}{qr} \]
- Substituting \( r = \frac{d}{2} \):
\[ B = \frac{mv}{q \cdot \frac{d}{2}} = \frac{2mv}{qd} \]

5. **Finding the Time Interval \( \Delta t \)**:
- The distance traveled by the charge while it is in the magnetic field is the circumference of the semicircle:
\[ \text{Circumference} = \pi r \]
- The distance traveled can also be expressed as:
\[ \text{Distance} = v \Delta t \]
- Setting these equal gives:
\[ v \Delta t = \pi r \]
- Substituting \( r = \frac{d}{2} \):
\[ v \Delta t = \pi \cdot \frac{d}{2} \]
- Solving for \( \Delta t \):
\[ \Delta t = \frac{\pi d}{2v} \]

### Final Results:
- The magnetic field \( B \) is given by:
\[ B = \frac{2mv}{qd} \]
- The time interval \( \Delta t \) is given by:
\[ \Delta t = \frac{\pi d}{2v} \]