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A circular current carrying coil has a radius \(R\). The distance from the centre of the coil on the axis where the magnetic induction will be \(\frac{1}{8}\) th of its value at the centre of the coil, is: (a) \(R \sqrt{3}\) (b) \(\frac{R}{\sqrt{3}}\) (c) \(\frac{2 R}{\sqrt{3}}\) (d) \((2 \sqrt{3}) R\)

Short Answer

Expert verified
The distance is \( R\sqrt{3} \).

Step by step solution

01

Understand the Magnetic Induction Formula

The magnetic induction at the center of a circular coil carrying current is given by the formula \( B = \frac{\mu_0 I}{2R} \), where \( \mu_0 \) is the permeability of free space, \( R \) is the radius, and \( I \) is the current flowing through the coil.
02

Express Magnetic Induction at a Point on the Axis

The magnetic induction \( B_x \) at a distance \( x \) from the center on the axis of the coil is given by the formula: \[ B_x = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \]
03

Set Induction on Axis to Fraction of Center Value

We are tasked to find \( x \) for which \( B_x = \frac{1}{8}B_0 \), where \( B_0 \) is the induction at the center: \[ \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} = \frac{1}{8} \times \frac{\mu_0 I}{2R} \]
04

Simplify and Solve the Equation

Simplifying, we have: \[ \frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R} \]Multiply both sides by \(8R(R^2 + x^2)^{3/2}\): \[ 8R^3 = (R^2 + x^2)^{3/2} \]Raise both sides to the power \(2/3\):\[ (8R^3)^{2/3} = R^2 + x^2 \]
05

Simplifying the Powers

Calculate \( \sqrt[3]{8R^3} \) which results in \( 2R \). Thus,\( (2R)^2 = R^2 + x^2 \) simplifies to:\( 4R^2 = R^2 + x^2 \).
06

Solve for x

Rearrange the equation:\[ 4R^2 - R^2 = x^2 \]\[ 3R^2 = x^2 \]Take the square root of both sides:\[ x = R\sqrt{3} \]
07

Select Correct Answer from Options

Evaluate the answer choices: option (a) \( R \sqrt{3} \) matches our solution.

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Key Concepts

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

Circular Current Carrying Coil
In the realm of electromagnetism, a circular current carrying coil is a simple yet fascinating object. Imagine a copper wire shaped into a perfect circle, through which an electric current flows. The uniform movement of electrons through the wire generates a magnetic field. This is a crucial principle in physics as it forms the basis of many technologies, like electromagnets and sensors.

The magnetic properties of the coil are shaped by various factors:
  • Radius of the Coil ( R) : The larger the radius, the more spread-out the magnetic field becomes.
  • Current ( I) : Stronger currents enhance magnetic fields.
  • Number of Turns: Inverted from our exercise, multiple loops of wire amplify the field.
This coil creates a concentrated magnetic field at its center, which fades as one moves outward and along the coil's axis. Understanding this behavior is key to managing and applying electromagnetic forces effectively in practical scenarios.
Magnetic Field on the Axis
A circular current carrying coil not only creates a magnetic field at its center but also extends this field along its axis. The field's strength diminishes as you move away from the center. Here's where physics and mathematics intersect beautifully.

When trying to determine the magnetic field on the axis of such a coil, a specific formula comes into play: \[ B_x = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \]This formula means that the field's strength (B_x) at any point along the axis depends on several variables:
  • The current (I) flowing through the coil.
  • The radius (\(R\)) of the coil.
  • The distance (\(x\)) from the coil's center along the axis.
Analyzing the equation, one can observe it reflects how the magnetic field decays with distance, controlled by the sum of the coil's radius and the axial distance. This subtle interaction between geometry and current is central to many magnetic applications.
Magnetic Field Formula
The magnetic field formula is not only a mathematical expression but a versatile tool that describes how a current influences its surrounding space. For a circular current carrying coil, this formula is simplified at its center as:\[ B = \frac{\mu_0 I}{2R} \] Here, the magnetic field (B) is at its maximum because all current loops contribute equally. To understand this:
  • \(\mu_0\): Known as the permeability of free space, it is a fundamental constant that describes how a magnetic field interacts with the vacuum.
  • Current (I): The more current, the stronger the field.
  • Radius (R): Inversely proportional to the magnetic field strength at the center.
Grasping this formula helps solve complex problems and sheds light on how electromagnetic principles govern technologies spanning from electrical motors to communication devices.
Physics Problem-Solving
Understanding physics is as much about applying concepts as it is about learning them. Physics problem-solving is a strategic approach where you identify, analyze, and mathematically solve problems using known principles. In our exercise, the goal was to find the distance along the axis of a coil where the magnetic field is a fraction of the field at the center.

The steps included:
  • Comprehension: Knowing the relationship between magnetic field strength and distance.
  • Application: Utilizing the formula \[B_x = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}\]
  • Simplification: Translating the condition \(B_x = \frac{1}{8}B_0\) into a solvable equation.
  • Calculation: Solving to find \(x\), the required distance.
The correct application of formulas is crucial. It transforms abstract theory into tangible solutions, enabling us to solve practical problems effortlessly. With practice, anyone can master the art of problem-solving in physics.

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