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A long rigid wire liens along the X - ax...

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

A

1260

B

`1280/3`

C

1310

D

420

Text Solution

AI Generated Solution

The correct Answer is:
To find the magnetic force on the segment of the wire between \( x = 1 \, \text{m} \) and \( x = 4 \, \text{m} \), we can follow these steps: ### Step 1: Identify the Given Information - Current \( I = 10 \, \text{A} \) (in the positive x-direction) - Magnetic field \( \vec{B} = \hat{i} + 2x^2 \hat{j} \) (in Tesla) - The segment of the wire is between \( x = 1 \, \text{m} \) and \( x = 4 \, \text{m} \). ### Step 2: Write the Expression for the Magnetic Force The magnetic force \( d\vec{F} \) on a small segment of wire \( d\vec{l} \) carrying current in a magnetic field is given by: \[ d\vec{F} = I \, d\vec{l} \times \vec{B} \] Here, \( d\vec{l} = dx \hat{i} \). ### Step 3: Substitute the Values Substituting \( d\vec{l} \) and \( \vec{B} \) into the equation: \[ d\vec{F} = I \, (dx \hat{i}) \times (\hat{i} + 2x^2 \hat{j}) \] Calculating the cross product: \[ d\vec{F} = 10 \, (dx \hat{i}) \times (\hat{i} + 2x^2 \hat{j}) \] Using the properties of cross products: \[ \hat{i} \times \hat{i} = 0 \quad \text{and} \quad \hat{i} \times \hat{j} = \hat{k} \] Thus, \[ d\vec{F} = 10 \, dx \, (0 + 2x^2 \hat{k}) = 20x^2 \, dx \, \hat{k} \] ### Step 4: Integrate to Find the Total Force To find the total force \( F \) on the segment from \( x = 1 \) to \( x = 4 \): \[ F = \int_{1}^{4} 20x^2 \, dx \] Calculating the integral: \[ F = 20 \int_{1}^{4} x^2 \, dx = 20 \left[ \frac{x^3}{3} \right]_{1}^{4} \] Calculating the limits: \[ = 20 \left( \frac{4^3}{3} - \frac{1^3}{3} \right) = 20 \left( \frac{64}{3} - \frac{1}{3} \right) = 20 \left( \frac{63}{3} \right) = 20 \times 21 = 420 \, \text{N} \] ### Final Answer The magnetic force on the segment of the wire between \( x = 1 \, \text{m} \) and \( x = 4 \, \text{m} \) is \( \boxed{420 \, \text{N}} \).
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