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.]
- \(\frac{7\mu_oI}{24}\)
- \(\frac{7\mu_oI}{12}\)
- \(\frac{\mu_oI}{8}\)
- \(\frac{\mu_oI}{6}\)
The Correct Option is A
Approach Solution - 1
To solve the problem, we need to find the magnitude of the line integral of the magnetic field \(\overrightarrow{B}\) along a straight line from the point \((- \sqrt{3}a, a, 0)\) to \((a, a, 0)\) caused by a current \(I\) along the +z direction.
1. Understanding the Magnetic Field due to an Infinite Wire:
The magnetic field at a distance \(r\) from a long straight current-carrying wire is given by Ampère's law:
\[
B = \frac{\mu_0 I}{2 \pi r}
\]
where \(\mu_0\) is the magnetic permeability of free space.
2. Geometry and Path of Integration:
The wire lies along the z-axis, so the magnetic field \(\overrightarrow{B}\) forms concentric circles around the wire in the xy-plane, with direction given by the right-hand rule.
The magnetic field at any point is tangential to a circle centered on the z-axis, and magnitude depends on the radial distance \(r = \sqrt{x^2 + y^2}\).
3. Parametrizing the Path:
The path is a straight line from \(\mathbf{r}_1 = (-\sqrt{3}a, a)\) to \(\mathbf{r}_2 = (a, a)\) in the xy-plane.
Since \(y\) is constant along the path (\(y=a\)), and \(x\) varies from \(-\sqrt{3}a\) to \(a\), the path vector differential is:
\[
d\mathbf{l} = dx\, \hat{i}
\]
4. Expression for \(\overrightarrow{B}\) and \(\overrightarrow{B} \cdot d\mathbf{l}\):
At a point \((x, a)\), the distance from the z-axis (wire) is:
\[
r = \sqrt{x^2 + a^2}
\]
The direction of \(\overrightarrow{B}\) at this point is tangential to the circle centered on the z-axis, which is perpendicular to the radius vector from the wire. The unit vector for \(\overrightarrow{B}\) is along \(\hat{\phi}\), perpendicular to \(\hat{r}\).
Thus, the magnetic field vector at \((x,a)\) in the xy-plane is:
\[
\overrightarrow{B} = \frac{\mu_0 I}{2 \pi r} \hat{\phi}
\]
Since \(\hat{\phi}\) at \((x,a)\) is given by:
\[
\hat{\phi} = \frac{-a \hat{i} + x \hat{j}}{r}
\]
The displacement vector differential along the path is along \(\hat{i}\), so:
\[
\overrightarrow{B} \cdot d\mathbf{l} = B_x dx = \left(\frac{\mu_0 I}{2 \pi r}\right) \left(\frac{-a}{r}\right) dx = -\frac{\mu_0 I a}{2 \pi r^2} dx
\]
5. Calculating the Integral:
\[
\int \overrightarrow{B} \cdot d\mathbf{l} = -\frac{\mu_0 I a}{2 \pi} \int_{x=-\sqrt{3}a}^{x=a} \frac{1}{x^2 + a^2} dx
\]
The integral of \( \frac{1}{x^2 + a^2} \) is:
\[
\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right)
\]
So:
\[
\int \overrightarrow{B} \cdot d\mathbf{l} = -\frac{\mu_0 I a}{2 \pi} \times \frac{1}{a} \left[\tan^{-1}\left(\frac{x}{a}\right)\right]_{-\sqrt{3}a}^{a} = -\frac{\mu_0 I}{2 \pi} \left[ \tan^{-1}(1) - \tan^{-1}(-\sqrt{3}) \right]
\]
Using values:
\[
\tan^{-1}(1) = \frac{\pi}{4}, \quad \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}
\]
So:
\[
\int \overrightarrow{B} \cdot d\mathbf{l} = -\frac{\mu_0 I}{2 \pi} \left( \frac{\pi}{4} + \frac{\pi}{3} \right) = -\frac{\mu_0 I}{2 \pi} \times \frac{7\pi}{12} = -\frac{7 \mu_0 I}{24}
\]
6. Magnitude of the Integral:
The problem asks for the magnitude, so:
\[
\left| \int \overrightarrow{B} \cdot d\mathbf{l} \right| = \frac{7 \mu_0 I}{24}
\]
Final Answer:
The magnitude of the line integral is \(\boxed{\frac{7 \mu_0 I}{24}}\).
Approach Solution -2
Useful formula: \[\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}\]
Since the path doesn’t enclose the wire, calculating \(\oint \vec{B} \cdot d\vec{l}\) requires determining the angle subtended by the path at the origin.
The points are \((-\sqrt{3}a, a)\) and \((a, a)\). In polar coordinates these points correspond to
\(r_1 = \sqrt{(-\sqrt{3}a)^2 + a^2} = 2a\), \(\theta_1 = \arctan{\frac{a}{-\sqrt{3}a}} = \arctan{-\frac{1}{\sqrt{3}}}= \frac{5\pi}{6}\) \(r_2 = \sqrt{(a)^2 + a^2} = \sqrt{2}a\), \(\theta_2 = \arctan{\frac{a}{a}} = \arctan{1}= \frac{\pi}{4}\)
The change in angle \(\Delta \theta = \frac{5\pi}{6} - \frac{\pi}{4} = \frac{10\pi - 3\pi}{12} = \frac{7\pi}{12}\)
The line integral is
\[\int \vec{B} \cdot d\vec{l} = \frac{\mu_0 I}{2\pi} \Delta \theta = \frac{\mu_0 I}{2\pi} \cdot \frac{7\pi}{12} = \frac{7\mu_0 I}{24}\]
Answer: (A) \(\frac{7\mu_0 I}{24}\)
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