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

Options

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

Answer

To solve the problem, we need to calculate the magnetic force on a segment of a wire carrying a current in a non-uniform magnetic field. Here are the steps to find the solution:

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

### Step 2: Write the expression for the magnetic force
The magnetic force \( \vec{F} \) on a current-carrying wire segment 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: Define the length vector
Since the wire lies along the x-axis, the differential length vector \( d\vec{L} \) can be expressed as:
\[ d\vec{L} = dx \hat{i} \]

### Step 4: Calculate the cross product \( d\vec{L} \times \vec{B} \)
Now we calculate the cross product:
\[ d\vec{L} \times \vec{B} = (dx \hat{i}) \times (\hat{i} + 2x^2 \hat{j}) \]
Using the distributive property of the cross product:
\[ = dx (\hat{i} \times \hat{i}) + dx (2x^2 \hat{i} \times \hat{j}) \]
Since \( \hat{i} \times \hat{i} = 0 \):
\[ = 0 + 2x^2 dx (\hat{k}) \]
Thus,
\[ d\vec{F} = I \cdot d\vec{L} \times \vec{B} = I \cdot 2x^2 dx \hat{k} \]

### Step 5: Integrate to find the total force
Now we integrate \( d\vec{F} \) from \( x = 1 \) to \( x = 4 \):
\[ \vec{F} = \int_{1}^{4} I \cdot 2x^2 \, dx \hat{k} \]
Substituting \( I = 10 \):
\[ \vec{F} = 10 \int_{1}^{4} 2x^2 \, dx \hat{k} \]
Calculating the integral:
\[ = 10 \cdot 2 \left[ \frac{x^3}{3} \right]_{1}^{4} \hat{k} \]
\[ = 20 \left( \frac{4^3}{3} - \frac{1^3}{3} \right) \hat{k} \]
\[ = 20 \left( \frac{64}{3} - \frac{1}{3} \right) \hat{k} \]
\[ = 20 \left( \frac{63}{3} \right) \hat{k} \]
\[ = 20 \cdot 21 \hat{k} \]
\[ = 420 \hat{k} \, \text{N} \]

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