A change q=1 C is at (3m,4m) and moving towards positive x-axis with constant velocity of 4 `m//s`. A long current carrying wire is at origin. Current in this wire is 2.A and towards positive z-axis. Magnetic force on the charge at given instant is

Options

A. `(1.64xx10^(7)hatK)N`
B. `-(1.92xx10^(-7)hatK)N`
C. `(4.8hatJ-3.2hatK)xx10^(-7)N`
D. `(1.6hatJ-4.8hatK)xx10^(-7)N`

Answer

To solve the problem, we need to calculate the magnetic force acting on a charge moving in the vicinity of a long current-carrying wire. Here’s how we can approach the solution step by step:

### Step 1: Identify the Given Values
- Charge \( q = 1 \, \text{C} \)
- Position of the charge \( (x, y) = (3 \, \text{m}, 4 \, \text{m}) \)
- Velocity of the charge \( v = 4 \, \text{m/s} \) (moving towards positive x-axis)
- Current in the wire \( I = 2 \, \text{A} \) (flowing in the positive z-direction)

### Step 2: Calculate the Distance from the Wire
The distance \( R \) from the wire (located at the origin) to the charge can be calculated using the Pythagorean theorem:
\[ R = \sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m} \]

### Step 3: Calculate the Magnetic Field \( B \) due to the Wire
The magnetic field \( B \) around a long straight current-carrying wire is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi R} \]
Where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \times 2}{2 \pi \times 5} = \frac{8 \times 10^{-7}}{10} = 8 \times 10^{-8} \, \text{T} \]
The direction of the magnetic field due to the right-hand rule is out of the plane (positive z-direction).

### Step 4: Determine the Direction of the Velocity and Magnetic Field
- Velocity \( \vec{v} \) is along the positive x-axis: \( \vec{v} = 4 \hat{i} \)
- Magnetic field \( \vec{B} \) is in the positive z-direction: \( \vec{B} = 8 \times 10^{-8} \hat{k} \)

### Step 5: Calculate the Magnetic Force \( \vec{F} \)
The magnetic force on a charge moving in a magnetic field is given by:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
Calculating the cross product:
\[ \vec{v} \times \vec{B} = (4 \hat{i}) \times (8 \times 10^{-8} \hat{k}) = 4 \times 8 \times 10^{-8} (\hat{i} \times \hat{k}) = 32 \times 10^{-8} (-\hat{j}) = -32 \times 10^{-8} \hat{j} \]
Thus, the force becomes:
\[ \vec{F} = 1 \times (-32 \times 10^{-8} \hat{j}) = -32 \times 10^{-8} \hat{j} \, \text{N} \]

### Step 6: Final Result
The magnetic force on the charge at the given instant is:
\[ \vec{F} = -3.2 \times 10^{-7} \hat{j} \, \text{N} \]
This indicates that the force is acting in the negative y-direction.