A particle having a mass of 0.5g carries a charge of `2.5xx10^-8C`. The particle is given an initial horizontal velocity of `6xx10^4ms^-1`. To keep the particle moving in a horizontal direction

Options

A. The magnetic field may be perpendicular to the direction of the velocity
B. the magnetic field should be along the direction of the velocity
C. magnetic field should have a minimum value of 3.27T
D. no magnetic field is required

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Convert the mass of the particle to kilograms
The mass of the particle is given as 0.5 g. To convert grams to kilograms, we use the conversion factor \(1 \text{ g} = 10^{-3} \text{ kg}\).

\[ \text{Mass (m)} = 0.5 \text{ g} = 0.5 \times 10^{-3} \text{ kg} = 0.0005 \text{ kg} \]

### Step 2: Identify the forces acting on the particle
The particle is subjected to two main forces:
1. The gravitational force (\(F_g\)) acting downwards, which can be calculated using the formula:
\[ F_g = m \cdot g \]
where \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).

2. The magnetic force (\(F_m\)) acting perpendicular to the velocity of the particle, which can be calculated using:
\[ F_m = Q \cdot V \cdot B \]
where \(Q\) is the charge of the particle, \(V\) is its velocity, and \(B\) is the magnetic field strength.

### Step 3: Set the forces equal for horizontal motion
For the particle to remain in horizontal motion, the magnetic force must balance the gravitational force:
\[ F_m = F_g \]
This gives us:
\[ Q \cdot V \cdot B = m \cdot g \]

### Step 4: Solve for the magnetic field strength \(B\)
Rearranging the equation gives us:
\[ B = \frac{m \cdot g}{Q \cdot V} \]

### Step 5: Substitute the known values
Now we can substitute the values into the equation:
- Mass \(m = 0.5 \times 10^{-3} \text{ kg}\)
- Charge \(Q = 2.5 \times 10^{-8} \text{ C}\)
- Velocity \(V = 6 \times 10^{4} \text{ m/s}\)
- Gravitational acceleration \(g = 9.8 \text{ m/s}^2\)

Substituting these values into the equation for \(B\):
\[ B = \frac{(0.5 \times 10^{-3} \text{ kg}) \cdot (9.8 \text{ m/s}^2)}{(2.5 \times 10^{-8} \text{ C}) \cdot (6 \times 10^{4} \text{ m/s})} \]

### Step 6: Calculate \(B\)
Calculating the numerator:
\[ 0.5 \times 10^{-3} \times 9.8 = 4.9 \times 10^{-3} \text{ kg m/s}^2 \]

Calculating the denominator:
\[ (2.5 \times 10^{-8}) \times (6 \times 10^{4}) = 1.5 \times 10^{-3} \text{ C m/s} \]

Now substituting these results back into the equation for \(B\):
\[ B = \frac{4.9 \times 10^{-3}}{1.5 \times 10^{-3}} = 3.27 \text{ T} \]

### Conclusion
The magnetic field strength required to keep the particle moving in a horizontal direction is approximately \(3.27 \, \text{T}\).