A long conducting wire is placed along Y-axis and a current of 5 A is flowing through it in positive Y-axis. A charged particle of charge ` + 3.2 xx 10^(-19)` C is moving parallel to the conductor towards negative Y-axis with a speed of `5 xx 10^(4) m s^(-1)`. Calculate the force exerted by the magnetic field of current on charged particle. The perpendicular distance between charged particle and wire is 25 cm.

Answer

To solve the problem, we need to calculate the magnetic force exerted on a charged particle moving in the magnetic field created by a long conducting wire carrying a current. Here are the steps to find the solution:

### Step 1: Identify the Given Values
- Current (I) = 5 A (flowing in the positive Y-direction)
- Charge of the particle (q) = \( +3.2 \times 10^{-19} \) C
- Speed of the charged particle (v) = \( 5 \times 10^4 \) m/s (moving in the negative Y-direction)
- Perpendicular distance (d) from the wire = 25 cm = 0.25 m

### Step 2: Calculate the Magnetic Field (B) due to the Wire
The magnetic field around a long straight conductor carrying current is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi d} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \times 5}{2 \pi \times 0.25} \]
\[ B = \frac{(4 \times 10^{-7}) \times 5}{0.5} = 4 \times 10^{-6} \, \text{T} \]

### Step 3: Determine the Direction of the Magnetic Field
Using the right-hand rule, if the current is flowing in the positive Y-direction, the magnetic field will circulate around the wire. At the position of the charged particle (which is below the wire), the magnetic field will be directed in the negative Z-direction.

### Step 4: Calculate the Force (F) on the Charged Particle
The magnetic force experienced by a charged particle moving in a magnetic field is given by:
\[ F = q(v \times B) \]
Since the charged particle is moving in the negative Y-direction and the magnetic field is in the negative Z-direction, we can use the right-hand rule to find the direction of the force.

1. The velocity vector \( \vec{v} = -v \hat{j} \)
2. The magnetic field vector \( \vec{B} = -B \hat{k} \)

Now, calculate the cross product:
\[ v \times B = (-v \hat{j}) \times (-B \hat{k}) = vB (\hat{j} \times \hat{k}) = vB \hat{i} \]

### Step 5: Substitute the Values into the Force Equation
Now substituting the values:
\[ F = q(vB) \]
Substituting \( B = 4 \times 10^{-6} \, \text{T} \):
\[ F = (3.2 \times 10^{-19}) \times (5 \times 10^4) \times (4 \times 10^{-6}) \]
Calculating:
\[ F = (3.2 \times 10^{-19}) \times (2 \times 10^{-1}) = 6.4 \times 10^{-20} \, \text{N} \]

### Step 6: Determine the Direction of the Force
The force is directed along the positive X-axis (as determined from the right-hand rule).

### Final Answer
The force exerted by the magnetic field on the charged particle is:
\[ F = 6.4 \times 10^{-20} \, \text{N} \quad \text{(in the positive X-direction)} \]

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