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5: When the current flowing in a circular coil is doubled and the number of turns of the coil in it is halved, the magnetic field at its centre will become (a) Four times (b) Same (c) Half (d) Double

Short Answer

Expert verified
The magnetic field at the center of the circular coil will become double when the current is doubled and the number of turns is halved (B2=2B1).

Step by step solution

01

Formula for Magnetic Field at the Center of a Circular Coil

The formula for the magnetic field (B) at the center of a circular coil with radius R, carrying a current I, and having n turns is given by: B=μ04π2πnIR Where μ0 is the permeability of free space. 2. Calculate the initial magnetic field
02

Initial Magnetic Field

Let's denote the initial magnetic field as B1. Using the formula, we have: B1=μ04π2πnIR 3. Apply the given changes to the coil, current, and number of turns
03

Changes in Current and Number of Turns

The problem states that the current flowing in the coil is doubled, and the number of turns of the coil is halved. So, we can write the new current as 2I, and the new number of turns as n2. 4. Calculate the new magnetic field
04

New Magnetic Field

Let's denote the new magnetic field as B2. Using the formula with the new current and number of turns, we have: B2=μ04π2π(n2)(2I)R 5. Compare the initial and new magnetic field
05

Comparison of Magnetic Fields

To find the relationship between the initial and new magnetic field, let's divide B2 by B1: B2B1=μ04π2π(n2)(2I)Rμ04π2πnIR=(n2)(2I)nI=2 6. Select the correct option
06

Correct Option

We found that the new magnetic field is two times the initial magnetic field (B2=2B1). Therefore, the correct answer is: (d) Double

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Key Concepts

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

Electromagnetism
Electromagnetism is a branch of physics that focuses on the study of electromagnetic forces. These forces arise due to the interaction between electrically charged particles. An essential aspect of electromagnetism is the way electric currents and magnetic fields relate to one another. When an electric current flows through a wire, it generates a magnetic field around it. This principle is vital when understanding applications like motors, transformers, and coils.

In the context of a circular coil, electromagnetism explains how the movement of electrons (current) through the coil produces a magnetic field at its center. The strength and direction of this magnetic field are influenced by factors such as the amount of current, the number of turns in the coil, and the coil's radius. The behavior of these elements is predictable by using specific formulas, helping us solve particular physics problems. By comprehending electromagnetism, students grasp the foundational theories driving many modern technologies.
Ampere's Law
Ampere's Law is a crucial equation in electromagnetism that links electric currents with magnetic fields. Named after the French physicist André-Marie Ampère, the law states that the integrated magnetic field around a closed loop is proportional to the electric current passing through it.

This can be mathematically expressed as:Bdl=μ0Iwhere:
  • B is the magnetic field,
  • dl is a differential element of the loop,
  • μ0 is the permeability of free space,
  • I is the net current through the loop.
In practical terms, Ampere's Law helps in calculating the strength of the magnetic field in different parts of a circuit. For a circular coil, we see how the current and turns affect the magnetic field using these principles. The law enriches our understanding by simplifying the often complex interactions between currents and magnetic fields.
Physics Problem Solving
Physics problem solving involves applying mathematical formulas and theories to understand and predict physical phenomena. It requires a combination of logical reasoning, theoretical knowledge, and the ability to interpret equations.

When solving a physics problem, like determining the magnetic field of a circular coil, follow a strategic approach:
  • Identify what is being asked and the relevant physical principles.
  • Write down known information and variables, such as current and number of turns.
  • Apply the appropriate formula, in this case, the one that links magnetic fields to currents and number of turns.
  • Perform calculations methodically to solve for the unknown quantity.
  • Analyze the results to ensure they make sense physically.
In this specific exercise, we saw how changes to current and coil turns affected the magnetic field. The adjustments indicate why understanding formulas and solving step-by-step is essential. By honing these skills, students can tackle a wide range of physics problems efficiently and accurately.

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Most popular questions from this chapter

A coil in the shape of an equilateral triangle of side 115 suspended between the pole pieces of a permanent magnet such that B is in plane of the coil. If due to a current I in the triangle a torque τ acts on it, the side 1 of the triangle is (a) (2/3)(τ/BI)1/2 (b) (2/3)(τ/BI) (c) 2[τ/(3)BI]1/2 (d) (1/3)(τ/BI)

1: A charged particle moves in a uniform mag. field. The velocity of the particle at some instant makes an acute angle with the mag. field. The path of the particle will be (a) A straight line (b) A circle (c) A helix with uniform pitch (d) A helix with non-uniform pitch

Two straight long conductors AOB and COD are perpendicular to each other and carry currents I1 and I2. The magnitude of the mag. field at a point " Pn at a distance " a from the point "O" in a direction perpendicular to the plane ABCD is (a) [(μ0)/(2πa)](I1+I2) (b) [(μ0)/(2πa)](I1I2) (c) [(μ0)/(2πa)](I12+I22)1/2 (d) [(μ0)/(2πa)][(I1I2)/(I1I2)]

The magnetic induction at a point P which is at a distance 4 cm from a long current carrying wire is 108 tesla. The field of induction at a distance 12 cm from the same current would be tesla. (a) 3.33×109 (b) 1.11×104 (c) 3×103 (d) 9×102

In the case of bar magnet, lines of magnetic induction (a) Start from the North pole and end at the South pole (b) Run continuously through the bar and outside (c) Emerge in circular paths from the middle of the bar (d) Are produced only at the North pole like rays of light from a bulb.

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