Question

An infinitely long wire, located on the z-axis, carries a current I along the +z-direction and produces the magnetic field $\bar{B}$ . The magnitude of the line integral $∫ \overrightarrow{B}.\overrightarrow{dl}$ along a straight line from the point $(−√3a, a, 0)$to $(a, a, 0)$ is given by [$\mu_0$ is the magnetic permeability of free space.]

• A. $\frac{7\mu_oI}{24}$
• B. $\frac{7\mu_oI}{12}$
• C. $\frac{\mu_oI}{8}$
• D. $\frac{\mu_oI}{6}$

Answer 1

The Correct Option is A

Magnetic Field and Line Integral 

1. Magnetic Field Due to a Current-Carrying Wire

The magnetic field at a distance $r$ from a current-carrying wire is given by:

$$\mathbf{B} = \frac{\mu_0 I}{2\pi r}$$

2. Line Integral Calculation

The line integral is given by:

$$\int \mathbf{B} \cdot d\mathbf{l} = \int B \cos\theta \, dl.$$

Geometry of the Path:

• The path extends from $(-\sqrt{3}a, a, 0)$ to $(a, a, 0)$.
• Parametric equation: $x \in [-\sqrt{3}a, a]$, $y = a$, $z = 0$.

The magnetic field at $(x, a, 0)$ due to the wire at the origin is:

$$\mathbf{B} = \frac{\mu_0 I}{2\pi \sqrt{x^2 + a^2}} \hat{\phi}$$

The element of the path is:

$$d\mathbf{l} = dx \hat{i}$$

The angle between $\mathbf{B}$ and $d\mathbf{l}$ is $\theta = \angle(\hat{\phi}, \hat{i})$, with:

$$\cos\theta = \frac{a}{\sqrt{x^2 + a^2}}.$$

3. Computing the Line Integral

Substituting into the integral:

$$\int \mathbf{B} \cdot d\mathbf{l} = \int_{-\sqrt{3}a}^{a} \frac{\mu_0 I}{2\pi \sqrt{x^2 + a^2}} \cdot \frac{a}{\sqrt{x^2 + a^2}} dx$$

$$= \frac{\mu_0 I a}{2\pi} \int_{-\sqrt{3}a}^{a} \frac{dx}{x^2 + a^2}.$$

4. Evaluating the Integral

Using the known integral result:

$$\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right),$$

we apply limits:

$$\int_{-\sqrt{3}a}^{a} \frac{dx}{x^2 + a^2} = \frac{1}{a} \left[ \tan^{-1} \left( \frac{a}{a} \right) - \tan^{-1} \left( \frac{-\sqrt{3}a}{a} \right) \right].$$

Simplifying:

$$= \frac{1}{a} \left( \tan^{-1}(1) - \tan^{-1}(-\sqrt{3}) \right).$$

Using standard values:

• $\tan^{-1}(1) = \frac{\pi}{4}$
• $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$

$$= \frac{1}{a} \left( \frac{\pi}{4} + \frac{\pi}{3} \right).$$

Calculating the sum:

$$= \frac{1}{a} \cdot \frac{7\pi}{12}.$$

5. Final Result

$$\int \mathbf{B} \cdot d\mathbf{l} = \frac{\mu_0 I a}{2\pi} \cdot \frac{1}{a} \cdot \frac{7\pi}{12} = \frac{7\mu_0 I}{24}.$$

Final Answer:

$$\frac{7\mu_0 I}{24}$$