A particle of mass `0.6g` and having charge of `25 n_(C)` is moving horizontally with a uniform velocity `1.2xx10^(4)ms^(-1)` in a uniform magnetic field, then the value of the magnetic induction is `(g=10ms^(-2))`

Options

A. Zero
B. 10T
C. 20T
D. 200T

Answer

To solve the problem, we need to find the value of the magnetic induction (B) when a charged particle is moving in a magnetic field. We can use the balance of forces acting on the particle.

### Step-by-Step Solution:

1. **Convert the given values to SI units:**
- Mass of the particle, \( m = 0.6 \, \text{g} = 0.6 \times 10^{-3} \, \text{kg} \)
- Charge of the particle, \( q = 25 \, \text{nC} = 25 \times 10^{-9} \, \text{C} \)
- Velocity of the particle, \( v = 1.2 \times 10^{4} \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the weight of the particle (W):**
\[ W = mg = (0.6 \times 10^{-3} \, \text{kg}) \times (10 \, \text{m/s}^2) = 6 \times 10^{-3} \, \text{N} \]

3. **Set up the equation for the magnetic force (F_m):**
The magnetic force acting on the charged particle can be expressed as:
\[ F_m = qvB \sin \theta \]
Since the particle is moving horizontally and the magnetic field is perpendicular to the velocity, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \). Therefore:
\[ F_m = qvB \]

4. **Equate the magnetic force to the weight of the particle:**
Since the magnetic force balances the weight of the particle:
\[ qvB = mg \]

5. **Rearranging the equation to solve for B:**
\[ B = \frac{mg}{qv} \]

6. **Substituting the known values into the equation:**
\[ B = \frac{(0.6 \times 10^{-3} \, \text{kg}) \times (10 \, \text{m/s}^2)}{(25 \times 10^{-9} \, \text{C}) \times (1.2 \times 10^{4} \, \text{m/s})} \]

7. **Calculating the numerator:**
\[ \text{Numerator} = 0.6 \times 10^{-3} \times 10 = 6 \times 10^{-3} \, \text{N} \]

8. **Calculating the denominator:**
\[ \text{Denominator} = (25 \times 10^{-9}) \times (1.2 \times 10^{4}) = 30 \times 10^{-5} = 3 \times 10^{-4} \, \text{C m/s} \]

9. **Final calculation for B:**
\[ B = \frac{6 \times 10^{-3}}{3 \times 10^{-4}} = 20 \, \text{T} \]

### Conclusion:
The value of the magnetic induction \( B \) is \( 20 \, \text{T} \).