A long rigid wire liens along the X - axis and carries a current of 10 A in the positive X - direction . Round the wire , the external magnetic field is `vec(B)= hati+2x^2hatj` with x in metres and B in Tesla. The magnetic force (in Sl units ) on the segment of the wire between x = 1m and x = 4 m is

Options

A. 1260
B. `1280/3`
C. 1310
D. 420

Answer

To solve the problem of finding the magnetic force on a segment of a wire carrying a current in a magnetic field, we can follow these steps:

### Step 1: Understand the Given Information
- Current \( I = 10 \, \text{A} \) in the positive X-direction.
- Magnetic field \( \vec{B} = \hat{i} + 2x^2 \hat{j} \) (where \( x \) is in meters and \( B \) is in Tesla).
- We need to find the magnetic force on the wire segment between \( x = 1 \, \text{m} \) and \( x = 4 \, \text{m} \).

### Step 2: Use the Formula for Magnetic Force
The magnetic force \( \vec{F} \) on a current-carrying wire in a magnetic field is given by:
\[ \vec{F} = I \int \vec{L} \times \vec{B} \, dx \]
where \( \vec{L} \) is the length vector of the wire segment.

### Step 3: Set Up the Element of the Wire
For a small segment of the wire \( dx \) at position \( x \), the length vector \( \vec{L} \) can be represented as:
\[ \vec{L} = dx \hat{i} \]

### Step 4: Calculate the Cross Product
Now, we need to calculate the cross product \( \vec{L} \times \vec{B} \):
\[ \vec{B} = \hat{i} + 2x^2 \hat{j} \]
Calculating the cross product:
\[ \vec{L} \times \vec{B} = (dx \hat{i}) \times (\hat{i} + 2x^2 \hat{j}) = dx (\hat{i} \times \hat{i} + 2x^2 \hat{i} \times \hat{j}) \]
Since \( \hat{i} \times \hat{i} = 0 \) and \( \hat{i} \times \hat{j} = \hat{k} \):
\[ \vec{L} \times \vec{B} = dx \cdot 2x^2 \hat{k} = 2x^2 dx \hat{k} \]

### Step 5: Substitute into the Force Equation
Substituting back into the force equation:
\[ \vec{F} = I \int_{1}^{4} (2x^2 dx \hat{k}) \]
Substituting \( I = 10 \):
\[ \vec{F} = 10 \int_{1}^{4} 2x^2 \, dx \hat{k} \]

### Step 6: Evaluate the Integral
Now we evaluate the integral:
\[ \int 2x^2 \, dx = \frac{2}{3} x^3 \]
Calculating the definite integral from 1 to 4:
\[ \int_{1}^{4} 2x^2 \, dx = \left[ \frac{2}{3} x^3 \right]_{1}^{4} = \frac{2}{3} (4^3 - 1^3) = \frac{2}{3} (64 - 1) = \frac{2}{3} \cdot 63 = 42 \]

### Step 7: Calculate the Total Force
Now substituting back into the force equation:
\[ \vec{F} = 10 \cdot 42 \hat{k} = 420 \hat{k} \, \text{N} \]

### Final Answer
The magnetic force on the segment of the wire between \( x = 1 \, \text{m} \) and \( x = 4 \, \text{m} \) is:
\[ \vec{F} = 420 \, \text{N} \, \hat{k} \]