/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A square frame of side \(1 \math... [FREE SOLUTION] | 91Ó°ÊÓ

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A square frame of side \(1 \mathrm{~m}\) carries a current \(i\), produces a magnetic field \(B\) at its centre. The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is \(B^{\prime}\). The ratio \(B / B^{\prime}\) is (a) \(\frac{8}{\pi^{2}}\) (b) \(\frac{8 \sqrt{2}}{\pi^{2}}\) (c) \(\frac{16}{\pi^{2}}\) (d) \(\frac{16}{\sqrt{2} \pi^{2}}\)

Short Answer

Expert verified
The ratio \( B / B^{\prime} \) is \( \frac{8 \sqrt{2}}{\pi^{2}} \). (Option b)

Step by step solution

01

Calculate the Perimeter of the Square

The square frame has a side length of 1m, so its perimeter is calculated as four times the side length: \[ P_{ ext{square}} = 4 imes 1 = 4 \ ext{meters}.\]
02

Determine the Radius of the Circular Coil

The circular coil has the same perimeter as the square, meaning that the circumference of the coil is 4 meters. Using the formula for the circumference of a circle \( C = 2\pi r \), we can solve for the radius \( r \):\[ 2\pi r = 4 \r = \frac{4}{2\pi} = \frac{2}{\pi} \ ext{meters}.\]
03

Calculate the Magnetic Field at the Center of the Square Frame

The magnetic field due to a current-carrying square loop at its center is given by:\[ B = \frac{2\sqrt{2}\mu_0 i}{a\pi}, \]where \( a \) is the side length of the square. Hence, for a side \( a = 1 \ ext{m} \), it becomes:\[ B = \frac{2\sqrt{2}\mu_0 i}{1\pi}.\]
04

Calculate the Magnetic Field at the Center of the Circular Coil

The magnetic field at the center of a circular coil is given by:\[ B^{\prime} = \frac{\mu_0 i}{2r}. \]Substitute the radius from Step 2:\[ B^{\prime} = \frac{\mu_0 i}{2 \times \frac{2}{\pi}} = \frac{\mu_0 i \pi}{4}.\]
05

Find the Ratio \( \frac{B}{B^{\prime}} \)

Now substitute the expressions for \( B \) and \( B^{\prime} \) from Steps 3 and 4:\[ \frac{B}{B^{\prime}} = \frac{\frac{2\sqrt{2}\mu_0 i}{\pi}}{\frac{\mu_0 i \pi}{4}}. \]Simplifying, this reduces to:\[ \frac{B}{B^{\prime}} = \frac{2\sqrt{2} \times 4}{\pi^2} = \frac{8\sqrt{2}}{\pi^2}.\]

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Key Concepts

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

Magnetic Field
A magnetic field is a region around a magnet, electric current, or changing electric field where magnetic forces can be detected. Magnetic fields are crucial for understanding electromagnetism. They are typically represented by the symbol \( B \) and measured in Tesla (T). The strength and direction of a magnetic field can influence various phenomena and are important in technology and natural processes.

When a current flows through a conductor, such as a wire, it creates a magnetic field around it. The right-hand rule helps visualize the field's direction: if you grip the wire with your right hand, your thumb points in the direction of the current, and your fingers curl in the direction of the magnetic field lines.

In more complex systems like loops or coils, the shape and current affect the field's characteristics. These configurations are foundational in many technologies, from motors to transformers.
Square and Circular Loop
To understand the differences in how magnetic fields behave with different loop shapes, consider a square loop and a circular loop. Both loops are fundamentally distinct in geometry, affecting how magnetic fields are generated within them.

A square loop with a particular side length has magnetic field lines that interact differently compared to a circular loop. In a square loop, the magnetic field at the center can be influenced by the symmetry and angles of the square, leading to a unique distribution of field lines.
  • The magnetic field due to a square loop is calculated using a more complex formula: \( B = \frac{2\sqrt{2}\mu_0 i}{a\pi} \).
  • This involves the side length and incorporates more geometric factors.

In contrast, a circular loop, with its smooth, symmetric shape, generates a magnetic field that is often easier to calculate. The formula for the magnetic field at its center is \( B^{\prime} = \frac{\mu_0 i}{2r} \). Here, \( r \) stands for the radius, showing how the geometric shape simplifies the equation.

The difference in calculation emphasizes how loop geometry affects magnetic field representation, making it an interesting study in physics where students learn to appreciate how shapes interact with forces.
Magnetic Field Ratio
The concept of comparing magnetic fields between different loop shapes is crucial in electromagnetism. By understanding magnetic field ratios, one can evaluate how geometry impacts magnetic force distribution.

Taking square and circular loops with equal perimeters, we can easily compare their magnetic fields. The ratio \( \frac{B}{B^{\prime}} \) gives insight into relative field strengths. From the given problem, and using the formulas:
  • The magnetic field center of a square is given by \( B = \frac{2\sqrt{2}\mu_0 i}{\pi} \).
  • The magnetic field of a circular coil is \( B^{\prime} = \frac{\mu_0 i \pi}{4} \).

To find the ratio \( \frac{B}{B^{\prime}} \), we substitute these values, leading to the simplification:
\[ \frac{B}{B^{\prime}} = \frac{8\sqrt{2}}{\pi^2} \]

This ratio demonstrates how the field strength contrasts based on shape, providing a clear quantitative expression for analysis and understanding the influence of geometric differences on magnetic fields in electric circuits.

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Most popular questions from this chapter

Consider a wire carrying a steady current, \(I\) placed in a uniform magnetic field \(\mathbf{B}\) perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that, \(\quad\) [NCERT Exemplar] (a) motion of charges inside the conductor is unaffected by B since they do not absorb energy (b) some charges inside the wire move to the surface as a result of B (c) if the wire moves under the influence of \(\mathrm{B}\), no work is done by the force (d) if the wire moves under the influence of \(\mathrm{B}\), no work is done by the magnetic force on the ions, assumed fixed within the wire

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