Question

Figure shows a thin strip of width b which carries a current I. Find magnetic induction due to the current in strip at point P located at a distance r from the strip as shown.

• A. The magnetic induction at point P due to the current in the strip can be found by considering the strip as a collection of infinitesimal current elements and integrating their contributions. Each infinitesimal element of width $dx$ at a distance $x$ from point P carries a current $dI = \frac{I}{b}dx$. The magnetic field $dB$ due to this element at point P is given by the Biot-Savart law. For a current element of length $dl$ carrying current $I$, $dB = \frac{\mu_0 I}{4\pi} \frac{dl \times \hat{r}}{r^2}$. In this case, $dl = dy$ (assuming the strip has a length along the y-axis, and the current flows in the y-direction), and $r$ is the distance from the element to P. However, the problem statement and the hint suggest a different approach where the width $b$ is in the horizontal direction and the current flows vertically.
• B. To find the magnetic induction at point P, we can consider the strip as an infinite sheet of current. However, the problem specifies a finite width $b$ and a distance $r$, implying a calculation for a finite conductor. The hint $dI = \frac{I}{b}dx$ suggests that the current is distributed uniformly across the width $b$. Let's assume the strip is oriented vertically, carrying current $I$ upwards. Point P is at a horizontal distance $r$ from the strip. We can divide the strip into infinitesimal vertical elements of width $dx$ and height $dy$. The current in such an element is $dI = \frac{I}{b}dx \frac{dy}{L}$ (assuming length $L$). This approach seems overly complicated given the problem statement.
• C. Let's consider the strip as a collection of infinitesimally thin wires, each of width $dx$ and carrying a current $dI = \frac{I}{b}dx$. Assume the strip is in the yz-plane, with its width along the x-axis from $x=0$ to $x=b$, and the current flows in the +z direction. Point P is located at $(r, 0, 0)$. For an infinitesimal wire element at position $x'$ along the x-axis, carrying current $dI$, the magnetic field at P can be calculated. However, the diagram suggests the strip is vertical and the current flows upwards, and P is at a horizontal distance $r$. Let's assume the strip is in the y-direction, from $y=0$ to $y=L$, and its width $b$ is in the x-direction, from $x=0$ to $x=b$. Point P is at $(r, y_p, 0)$. This interpretation also seems complex.
• D. Let's consider the strip as a collection of infinitesimal current elements. Assume the strip is oriented vertically, and its width $b$ is along the x-axis. Let the strip occupy the region $0 \le x' \le b$ and extend infinitely along the y-axis. The current $I$ flows in the +y direction. Point P is at a distance $r$ from the strip, say at $(r, 0, 0)$. An infinitesimal element of width $dx'$ at position $x'$ carries a current $dI = \frac{I}{b}dx'$. This element can be treated as a thin wire of length $L$ (if finite) or infinitely long. The magnetic field $dB$ at P due to this element is given by Ampere's law for a long straight wire, $dB = \frac{\mu_0 dI}{2\pi r'}$, where $r'$ is the distance from the wire element to P. Here, $r' = \sqrt{x'^2 + y^2}$ if P is at $(r, y, 0)$ and the strip is at $x'=0$. However, the diagram shows P at a distance $r$ from the strip, and $x$ as the horizontal distance from a point on the strip to P. This implies the strip has width $b$ in the horizontal direction, and the current flows perpendicular to the plane of the figure (say, out of the page). Let's assume the strip is in the xy-plane, with width $b$ along the x-axis, from $x=0$ to $x=b$. The current flows in the +z direction. Point P is at $(r, 0, 0)$. An infinitesimal element of width $dx$ at position $x'$ carries current $dI = \frac{I}{b}dx$. This element is a wire of length $L$ (if the strip has a length $L$). The distance from this element to P is $\sqrt{r^2 + x'^2}$. The magnetic field $dB$ at P due to this element is $dB = \frac{\mu_0 dI}{2\pi \sqrt{r^2 + x'^2}}$. To find the total field, we integrate from $x'=0$ to $x'=b$: $B = \int_{0}^{b} \frac{\mu_0}{2\pi} \frac{I}{b} \frac{dx'}{\sqrt{r^2 + x'^2}}$. This integral evaluates to $B = \frac{\mu_0 I}{2\pi b} \left[ \ln(x' + \sqrt{r^2 + x'^2}) \right]_0^b = \frac{\mu_0 I}{2\pi b} \ln\left(\frac{b + \sqrt{r^2 + b^2}}{r}\right)$.

Answer 1

Answer: (D)

The magnetic induction at point P due to the current in the strip is given by $B = \frac{\mu_0 I}{2\pi b} \ln\left(\frac{b + \sqrt{r^2 + b^2}}{r}\right)$.

Answer 2

To find the magnetic induction at point P, we can consider the thin strip as a collection of infinitesimally thin current-carrying wires, each of width $dx$. The total current $I$ is distributed uniformly across the width $b$, so the current in an infinitesimal element of width $dx$ is $dI = \frac{I}{b}dx$.

Let's assume the strip lies in the xy-plane, with its width extending along the x-axis from $x=0$ to $x=b$. The current flows in the +z direction (perpendicular to the plane of the figure). Point P is located at a horizontal distance $r$ from the strip, which we can place at $(r, 0, 0)$ in Cartesian coordinates.

Each infinitesimal element of width $dx$ at position $x'$ along the x-axis can be treated as a thin, straight wire carrying current $dI$. The distance from this wire element to point P is $R = \sqrt{r^2 + x'^2}$.

According to the Biot-Savart law, the magnetic field $dB$ at point P due to this infinitesimal current element $dI$ is given by: $dB = \frac{\mu_0 dI}{2\pi R}$

Substituting $dI = \frac{I}{b}dx$ and $R = \sqrt{r^2 + x'^2}$: $dB = \frac{\mu_0}{2\pi} \frac{\frac{I}{b}dx}{\sqrt{r^2 + x'^2}} = \frac{\mu_0 I}{2\pi b} \frac{dx}{\sqrt{r^2 + x'^2}}$

To find the total magnetic induction $B$ at point P, we need to integrate $dB$ over the entire width of the strip, from $x'=0$ to $x'=b$: $B = \int_{0}^{b} dB = \int_{0}^{b} \frac{\mu_0 I}{2\pi b} \frac{dx}{\sqrt{r^2 + x^2}}$

The integral $\int \frac{dx}{\sqrt{r^2 + x^2}}$ is a standard integral, which evaluates to $\ln(x + \sqrt{r^2 + x^2})$.

So, evaluating the definite integral: $B = \frac{\mu_0 I}{2\pi b} \left[ \ln(x + \sqrt{r^2 + x^2}) \right]_{0}^{b}$ $B = \frac{\mu_0 I}{2\pi b} \left( \ln(b + \sqrt{r^2 + b^2}) - \ln(0 + \sqrt{r^2 + 0^2}) \right)$ $B = \frac{\mu_0 I}{2\pi b} \left( \ln(b + \sqrt{r^2 + b^2}) - \ln(r) \right)$ $B = \frac{\mu_0 I}{2\pi b} \ln\left(\frac{b + \sqrt{r^2 + b^2}}{r}\right)$

The direction of the magnetic field can be determined using the right-hand rule. If the current is flowing out of the page (+z direction) and P is to the right (in the +x direction), the magnetic field will be in the +y direction (into the page, if we consider a standard coordinate system where P is on the x-axis and the strip is on the y-axis, with current in z). However, the question asks for the magnitude of the magnetic induction.