Chapter 20: Problem 20
A circular coil has 100 turns and a mean diameter of \(10 \mathrm{~cm}\). It carries a current of \(2 \mathrm{~A}\). The strength of the magnetic field at a point on its axis at a distance \(2 \mathrm{~m}\) from the centre of the coil is approximately (a) \(6.28 \times 10^{-7} \mathrm{~Wb} / \mathrm{m}^{2}\) (b) \(3.14 \times 10^{-7} \mathrm{~Wb} / \mathrm{m}^{2}\) (c) \(1.57 \times 10^{-7} \mathrm{~Wb} / \mathrm{m}^{2}\) (d) \(10^{-7} \mathrm{~Wb} / \mathrm{m}^{2}\)
Short Answer
Step by step solution
Determine the Radius of the Coil
Recall the Formula for Magnetic Field on the Axis of a Coil
Substitute Known Values into the Formula
Simplify the Denominator and Numerator
Calculate the Magnetic Field
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Circular Coil
Permeability of Free Space
- \(\mu_0\) is part of the magnetic field formula, linking current and distance to magnetic field strength.
- It ensures that units work within calculations, as it is measured in teslas per ampere per meter.
- The permeability of free space is a universal constant and is the same everywhere, making it a reliable base for calculations in vacuums.
Magnetic Field Formula
- \(B\) is the magnetic field strength we want to calculate.
- \(\mu_0\) is the permeability of free space, providing the ground factor for the equation.
- \(N\) represents the number of turns in the coil, amplifying the magnetic field.
- \(I\) is the current in amperes flowing through the coil.
- \(r\) is the radius of the coil.
- \(x\) is the distance from the coil to the point where the field is measured.
Current in a Coil
- More current results in a stronger magnetic field. An increase in current increases the field intensively.
- In practical applications, managing current levels is crucial for controlling the magnetic field strength to reach desired outcomes.
- Within the magnetic field formula, this current is multiplied by other elements, amplifying the field produced by the coil.