Question

A long straight conductor is bent at an angle of 90° as shown in the figure. Magnetic field induction at the point A is

• A. $\frac{\mu_0}{4\pi}\frac{I}{h}[\frac{\sqrt{2}+1}{\sqrt{2}}]$
• B. $\frac{\mu_0}{4\pi}\frac{2I}{h}[\frac{\sqrt{2}+1}{\sqrt{2}}]$
• C. $\frac{\mu_0}{4\pi}\frac{I}{h}[\frac{\sqrt{2}+1}{2\sqrt{2}}]$
• D. $\frac{\mu_0}{4\pi}\frac{2I}{h}[\frac{\sqrt{3}+1}{\sqrt{3}}]$

Answer 1

Answer: (B)

(B)

Answer 2

The conductor consists of two semi-infinite straight wires meeting at a 90° angle. We assume the current flows such that the magnetic fields produced by each segment at point A add up, which occurs if point A is in the "inner" region of the bend. Let the corner be at the origin (0,0). Let the horizontal wire extend from $(-\infty, 0)$ to $(0,0)$ with current in the +x direction, and the vertical wire extend from $(0,0)$ to $(0, +\infty)$ with current in the +y direction. For the fields to add up at point A, A must be in the second quadrant, e.g., at $(-h, h)$.

1. Magnetic field due to the horizontal wire (Wire 1): The wire extends from $x = -\infty$ to $x = 0$ along the x-axis, with current $I$ flowing in the +x direction. Point A is at $(-h, h)$. The perpendicular distance from A to Wire 1 is $d_1 = h$. The magnetic field due to a semi-infinite wire is given by $B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)$. Here, $\theta_1$ is the angle made by the line joining A to the finite end $(0,0)$ with the perpendicular from A to the wire (at $(-h,0)$). This angle is $45^\circ$. $\theta_2$ is the angle made by the line joining A to the infinite end $(-\infty, 0)$ with the perpendicular. This angle is $90^\circ$. So, $B_1 = \frac{\mu_0 I}{4\pi h} (\sin 45^\circ + \sin 90^\circ) = \frac{\mu_0 I}{4\pi h} \left(\frac{1}{\sqrt{2}} + 1\right)$. Using the right-hand thumb rule, for current in +x direction and point A above the wire, the magnetic field $B_1$ is directed out of the page.

2. Magnetic field due to the vertical wire (Wire 2): The wire extends from $y = 0$ to $y = +\infty$ along the y-axis, with current $I$ flowing in the +y direction. Point A is at $(-h, h)$. The perpendicular distance from A to Wire 2 is $d_2 = h$. $\theta_1$ is the angle made by the line joining A to the finite end $(0,0)$ with the perpendicular from A to the wire (at $(0,h)$). This angle is $45^\circ$. $\theta_2$ is the angle made by the line joining A to the infinite end $(0, +\infty)$ with the perpendicular. This angle is $90^\circ$. So, $B_2 = \frac{\mu_0 I}{4\pi h} (\sin 45^\circ + \sin 90^\circ) = \frac{\mu_0 I}{4\pi h} \left(\frac{1}{\sqrt{2}} + 1\right)$. Using the right-hand thumb rule, for current in +y direction and point A to the left of the wire, the magnetic field $B_2$ is directed out of the page.

3. Total Magnetic Field: Since both $B_1$ and $B_2$ are in the same direction (out of the page), their magnitudes add up: $B_{net} = B_1 + B_2 = \frac{\mu_0 I}{4\pi h} \left(\frac{1}{\sqrt{2}} + 1\right) + \frac{\mu_0 I}{4\pi h} \left(\frac{1}{\sqrt{2}} + 1\right)$ $B_{net} = \frac{\mu_0 I}{4\pi h} \left(2 \left(\frac{1}{\sqrt{2}} + 1\right)\right) = \frac{\mu_0 I}{4\pi h} \left(\frac{2}{\sqrt{2}} + 2\right)$ $B_{net} = \frac{\mu_0 I}{4\pi h} (\sqrt{2} + 2) = \frac{\mu_0}{4\pi} \frac{2I}{h} \left(\frac{\sqrt{2} + 1}{\sqrt{2}}\right)$.