A wire lying along y-axis from `y=0` to `y=1 m` carries a current of `2mA` in the negative y-direction. The wire lies in a non uniform magnetic fiedl given by `B=(0.3T/m)yhati +(0.4Tm)yhatj`. The magnetic force on the entire wire is

Options

A. `-3xx10^-4hatjN`
B. `6xx10^-3hatkN`
C. `-3xx10^-4hatkN`
D. `3xx10^-4hatkN`

Explanation

To solve the problem of finding the magnetic force on a wire lying along the y-axis in a non-uniform magnetic field, we can follow these steps:

### Step 1: Understand the Given Information
- The wire extends from \( y = 0 \) to \( y = 1 \) meter.
- The current \( I = 2 \, \text{mA} = 2 \times 10^{-3} \, \text{A} \) flows in the negative y-direction.
- The magnetic field is given by \( \mathbf{B} = (0.3 \, \text{T/m})y \hat{i} + (0.4 \, \text{T}) \hat{j} \).

### Step 2: Identify the Magnetic Force Formula
The magnetic force \( \mathbf{F} \) on a current-carrying wire in a magnetic field is given by:
\[ \mathbf{F} = I \int \mathbf{B} \times d\mathbf{l} \]
where \( d\mathbf{l} \) is the differential length vector along the wire.

### Step 3: Set Up the Differential Length Element
Since the wire is along the y-axis, the differential length element can be expressed as:
\[ d\mathbf{l} = dy \hat{j} \]

### Step 4: Calculate the Magnetic Field at Each Point
The magnetic field \( \mathbf{B} \) can be evaluated at any point along the wire. Since \( y \) varies from 0 to 1, we have:
\[ \mathbf{B} = (0.3y) \hat{i} + (0.4) \hat{j} \]

### Step 5: Calculate the Cross Product
Now, we need to calculate the cross product \( \mathbf{B} \times d\mathbf{l} \):
\[ \mathbf{B} \times d\mathbf{l} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.3y & 0.4 & 0 \\ 0 & 1 & 0 \end{vmatrix} \]
Calculating this determinant gives:
\[ \mathbf{B} \times d\mathbf{l} = \hat{k}(0.3y \cdot 1 - 0.4 \cdot 0) = 0.3y \hat{k} \]

### Step 6: Substitute into the Force Equation
Now substitute this into the force equation:
\[ \mathbf{F} = I \int_0^1 (0.3y \hat{k}) \, dy \]
This simplifies to:
\[ \mathbf{F} = I \hat{k} \int_0^1 0.3y \, dy \]

### Step 7: Perform the Integration
Now, perform the integration:
\[ \int_0^1 0.3y \, dy = 0.3 \left[ \frac{y^2}{2} \right]_0^1 = 0.3 \cdot \frac{1^2}{2} = 0.15 \]

### Step 8: Calculate the Total Force
Now substitute back into the force equation:
\[ \mathbf{F} = I \hat{k} (0.15) = (2 \times 10^{-3} \, \text{A}) \hat{k} (0.15) = 3 \times 10^{-4} \hat{k} \, \text{N} \]

### Step 9: Final Result
Thus, the magnetic force on the entire wire is:
\[ \mathbf{F} = 3 \times 10^{-4} \hat{k} \, \text{N} \]