The magnetic field in a region is given by `B = B_(0) (1 - (X)/(L)) K`. A straight conductor AB of length 2L, placed with its ends at points A (a,0,0) and B(a,2L,0), carries a current I from A to B. Find the magnetic force on the conductor

Answer

To find the magnetic force on the conductor AB, we will follow these steps:

### Step 1: Understand the Magnetic Field
The magnetic field is given by:
\[ \mathbf{B} = B_0 \left(1 - \frac{X}{L}\right) \hat{k} \]
This indicates that the magnetic field varies with the x-coordinate and is directed along the z-axis (k-direction).

### Step 2: Identify the Coordinates of Points A and B
The coordinates of the points are:
- Point A: \( (a, 0, 0) \)
- Point B: \( (a, 2L, 0) \)

The length of the conductor AB is \( 2L \) and it is oriented along the y-axis.

### Step 3: Calculate the Magnetic Field at Point A
Since the magnetic field varies with the x-coordinate, we need to evaluate it at point A where \( X = a \):
\[ \mathbf{B}(a) = B_0 \left(1 - \frac{a}{L}\right) \hat{k} \]

### Step 4: Define the Length Vector of the Conductor
The conductor AB has a length vector \( \mathbf{L} \) which is directed along the y-axis:
\[ \mathbf{L} = 2L \hat{j} \]

### Step 5: Use the Formula for Magnetic Force
The magnetic force \( \mathbf{F} \) on a current-carrying conductor is given by:
\[ \mathbf{F} = I \mathbf{L} \times \mathbf{B} \]
Substituting the values we have:
\[ \mathbf{F} = I (2L \hat{j}) \times \left(B_0 \left(1 - \frac{a}{L}\right) \hat{k}\right) \]

### Step 6: Calculate the Cross Product
To compute the cross product \( \hat{j} \times \hat{k} \):
\[ \hat{j} \times \hat{k} = \hat{i} \]
Thus, we can simplify the force:
\[ \mathbf{F} = I (2L) B_0 \left(1 - \frac{a}{L}\right) \hat{i} \]

### Step 7: Final Expression for the Magnetic Force
The final expression for the magnetic force on the conductor is:
\[ \mathbf{F} = 2I B_0 \left(1 - \frac{a}{L}\right) L \hat{i} \]

### Summary
The magnetic force on the conductor AB is directed along the x-axis and is given by:
\[ \mathbf{F} = 2I B_0 \left(1 - \frac{a}{L}\right) L \hat{i} \]