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Chapter 19: Problem 23

A steady current \(i\) flows in a small square loop of wire of side \(l\) in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let \(M_{1}\) and \(M_{2}\) respectively denote the magnetic moments due to current loop before and after folding. Then (a) \(M_{2}=0\) (b) \(M_{1}\) and \(M_{2}\) are in the same direction (c) \(M_{1} / M_{2}=\sqrt{2}\) (d) \(M_{1} / M_{2}=1 / \sqrt{2}\)

Short Answer

Expert verified
Option (d) is correct: \( M_1 / M_2 = \sqrt{2} \).

Step by step solution

01

Calculate Initial Magnetic Moment

The magnetic moment of a current-carrying loop is given by \(M = I \times A \), where \(I\) is the current and \(A\) is the area of the loop. The area of the square loop with side length \(l\) is \(l^2\). Therefore, the initial magnetic moment \(M_1 = i \times l^2\).
02

Analyze the Configuration After Folding

After folding the square loop, it forms two triangular loops. Each side of the square now forms the hypotenuse of the triangle with sides in the vertical and horizontal planes.
03

Calculate Magnetic Moment After Folding

For each triangular loop, a perpendicular segment is needed to calculate area contributions to magnetic moment. Each triangle has sides of \(\frac{l}{2}\), and segments perpendicular are along the vertical plane (\(\frac{l}{2}\)) and horizontal plane (\(\frac{l}{2}\)). The effective area for each triangle becomes \(\frac{1}{2} \times \frac{l}{2} \times \frac{l}{2} = \frac{l^2}{8}\).
04

Compute Total Magnetic Moment (M2)

Both triangles contribute equally to the total magnetic moment. The effective area being \(\frac{l^2}{8}\) leads to an effective magnetic moment for each triangle as \(i \times \frac{l^2}{8}\). Since both triangles contribute, total magnetic moment \(M_2 = 2 \times i \times \frac{l^2}{8} = \frac{i l^2}{4}\).
05

Compare M1 and M2

The new magnetic moment \(M_2 = \frac{i l^2}{4}\) should be compared with the original \(M_1 = i l^2\). This gives \(\frac{M_1}{M_2} = \frac{i l^2}{\frac{i l^2}{4}} = 4\).
06

Analyze Direction of M1 and M2

The direction of the magnetic moment is given by the right-hand rule which remains in the same overall direction for the folded and original loop, as the areas oppose each other in equal parts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

current-carrying loop
A current-carrying loop is a closed loop of wire through which an electric current flows. This setup creates a magnetic field around it, resulting in a magnetic moment. The magnetic moment is a vector property that quantifies the magnetic influence of the current.
It is calculated using the formula:
  • \( M = I \times A \), where \( M \) is the magnetic moment, \( I \) is the current, and \( A \) is the area of the loop.
In the given exercise, the loop is initially a square with a side \( l \), making the area \( A = l^2 \).
Thus, the initial magnetic moment is \( M_1 = i \times l^2 \).
The alignment of the loop's area plays a critical role in determining the direction and magnitude of the magnetic moment. Understanding how this area changes when the loop is folded is crucial for calculating any new magnetic moments.
right-hand rule
The right-hand rule is a simple technique used to determine the direction of the magnetic moment in a current-carrying loop. By using your right hand, you can visualize the direction:
  • Point your thumb in the direction of the current flow.
  • Your curled fingers around the loop will naturally point in the direction of the magnetic moment.

In the context of the exercise, before and after folding the loop, we use the right-hand rule to confirm the direction of the magnetic moment.
Even after folding, if you apply the right-hand rule to both halves of the loop, it will show that the magnetic moment is still aligned in the same general direction as the original configuration. This helps clarify that despite changes in the loop's shape, the magnetic influence direction remains constant.
area of loop
Understanding the area of the loop is key to calculating the magnetic moment. Initially, the square loop has an area of \( l^2 \) when fully extended. When folded, the loop splits into two triangular parts, each with reduced, effective areas that contribute to the new magnetic moment.
When folded, each triangular section has a base and height of \( \frac{l}{2} \), calculating this gives:
  • Area of a triangle = \( \frac{1}{2} \times \frac{l}{2} \times \frac{l}{2} = \frac{l^2}{8} \).
Since there are two triangles contributing equally to the magnetic moment, the total effective area becomes \( \frac{l^2}{4} \).
This new area reflects how folding alters the loop's geometry, which directly influences the magnetic moment evaluation \( M_2 \). These area calculations ensure precise determination of the loop's magnetic behavior in varied physical configurations.

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Most popular questions from this chapter

A length \(l\) of wire carries a steady current \(i\). It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give three loops of smaller radius. The magnetic field at the centre caused by the same current is (a) one-third of its value (b) unaltered (c) three times of its initial value (d) nine times of its initial value

A galvanometer of resistance \(20 \Omega\) shown deflection of 10 divisions. When a current of \(1 \mathrm{~mA}\) is passed through it. If a shunt of \(4 \Omega\) is connected and there are 50 divisions on the scale, the range of the galvanometer is [Kerala CET 2007] (a) \(1 \mathrm{~A}\) (b) \(3 \mathrm{~A}\) (c) \(10 \mathrm{~mA}\) (d) \(30 \mathrm{~mA}\) (e) \(11 \mathrm{~mA}\)

Two insulating plates are both uniformly charged in such a way that the potential difference between them is \(V_{2}-V_{1}=20 \mathrm{~V}\) (i.e, plate 2 is at a higher potential). The plates are separated by \(d=0.1 \mathrm{~m}\) and can be treated as infinitely large. An electron is released from rest on the inner surface of plate \(1 .\) What is its speed when it hits plate 2 ? [AIEEE 2006] (a) \(32 \times 10^{-19} \mathrm{~ms}^{-1}\) (b) \(2.65 \times 10^{6} \mathrm{~ms}^{-1}\) (c) \(7.02 \times 10^{12} \mathrm{~ms}^{-1}\) (d) \(1.87 \times 10^{6} \mathrm{~ms}^{-1}\)

Assertion A force of \(1 \mathrm{~kg}\)-wt acts on \(1 \mathrm{~m}\) long wire carrying 10 A current held at \(90^{\circ}\) to a magnetic field of \(0.98 \mathrm{~T}\). Reason \(F=B i l \sin \theta\)

Match the following of Column I with Column II. Column I Column II I. Lorentz force A. \(\oint E \cdot \mathrm{d} \mathbf{A}=\frac{q}{\varepsilon_{0}}\) II. Gauss's law B. \(\quad \mathbf{d B}=\frac{\mu_{0}}{4 \pi} \frac{i \mathbf{d} l \times \mathbf{r}}{r^{3}}\) III. Biot-Savart law C. \(\quad \mathbf{F}=q(\mathbf{E}+(\mathbf{v} \times \mathbf{B}))\) IV. Coulomb's Law D. \(\quad F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)
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