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Chapter 13: Problem 1966

The effective length of a magnet is \(31.4 \mathrm{~cm}\) and its pole strength is \(0.5 \mathrm{~A} \mathrm{~m}\). The magnetic moment, if it is bent in the form of a semicircle will be Amp.m \(^{2}\) (a) \(0.1\) (b) \(0.01\) (c) \(0.2\) (d) \(1.2\)

Short Answer

Expert verified
The magnetic moment of a magnet bent in the form of a semicircle with an effective length of \(31.4 \mathrm{~cm}\) and a pole strength of \(0.5 \mathrm{~A} \mathrm{~m}\) is \(0.1 \mathrm{~A} \mathrm{~m}^2\).

Step by step solution

01

Identify the given values

We are given the following information: - Effective length of the magnet = 31.4 cm - Pole strength of the magnet = 0.5 A m
02

Convert the effective length to meters

To carry out our calculations, we must convert the given effective length from cm to meters: Effective length of the magnet = 31.4 cm × (1 m / 100 cm) = 0.314 m
03

Calculate the diameter of the circle formed by the semicircle

Since the magnet is bent in the form of a semicircle, the effective length would represent half of the circumference of the formed circle. To find the diameter of the circle, we can use the formula for the circumference: Circumference = 2 × π × radius Let the radius of the circle be R. Then, the effective length equals half the circumference: 0.314 m = π × R Now, we can solve for R: R = 0.314 m / π ≈ 0.1 m The diameter of the circle, which represents the distance between the poles, is twice the radius: Diameter = 2 × R = 2 × 0.1 m = 0.2 m
04

Calculate the magnetic moment

Now that we know the distance between the poles (the diameter of the circle), we can calculate the magnetic moment using the formula: Magnetic moment = pole strength × distance between poles Magnetic moment = 0.5 A m × 0.2 m = 0.1 A m²
05

Select the correct option

The calculated magnetic moment is 0.1 A m². Hence, the correct option is: (a) 0.1

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Key Concepts

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

Pole Strength
Pole strength refers to the ability of a magnetic pole to produce a magnetic field. It is measured in units of Amperes-meter (A m).
In simpler terms, it is a measure of the "magnetization" capability of a magnet's pole. Imagine pole strength like how strongly a magnet can "pull" at another magnetic object.
It is crucial when calculating the magnetic moment, as it determines part of the force with which the magnet can interact with others or influence its surroundings. Magnetic pole strength is consistent no matter the shape of the magnet, whether it is straight or bent, such as in a semicircle form.
When you know the pole strength, calculating the magnetic moment becomes a straightforward task, involving multiplication with the effective length, or distance between the poles.
Effective Length of Magnet
The effective length of a magnet is the distance over which the magnetic force acts within the magnet. In other words, it is the separation between the two poles of a magnet; where one end is the north pole, and the other is the south pole.
When a magnet is bent into a different shape, such as a semicircle, the effective length isn't simply the straight length of the magnet. The curved path of a magnet in a semicircular shape demands a different approach.
For a semicircle, the effective length is considered as half the circumference of the full circle that would be formed. Thus, a magnet bent into a semicircle with an effective length of 0.314 meters can help find the diameter of the full circle by solving the equation: \[\text{Effective length} = \frac{2\pi R}{2},\] simplifying to find the radius and then the diameter from the radius. This helps in further calculations of properties like the magnetic moment.
Semicircle
A semicircle, by definition, is half of a circle. When a magnet is shaped into a semicircle, its magnetic attributes don't disappear or lessen; they simply rearrange according to this new shape.
In this arrangement, the semicircle becomes important because it represents the geometry that dictates how magnetic forces will spread and interact. Understanding this rearrangement is crucial for precise calculations of related properties like the magnetic moment.
To find properties like the magnetic moment when dealing with a semicircle, we need to understand that effective length, or the straight-line distance between the poles, becomes the semicircular path effectively half the circle's circumference. Calculating this involves aligning the effective length with this path to determine the diameter, and from there, derive values needed for more complex calculations.
  • Effective length contributes directly to understanding the diameter.
  • Semicircles create unique magnetic conditions due to curvature.
Knowing this can then help properly determine how properties like pole strength affect this new geometrical arrangement.

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Most popular questions from this chapter

At a distance of \(10 \mathrm{~cm}\) from a long straight wire carrying current, the magnetic field is \(4 \times 10^{-2}\). At the distance of \(40 \mathrm{~cm}\), the magnetic field will be Tesla. (a) \(1 \times 10^{-2}\) (b) \(2 \times \overline{10^{-2}}\) (c) \(8 \times 10^{-2}\) (d) \(16 \times 10^{-2}\)

A small bar magnet has a magnetic moment \(1.2 \mathrm{~A} \cdot \mathrm{m}^{2}\). The magnetic field at a distance \(0.1 \mathrm{~m}\) on its axis will be tesla. (a) \(1.2 \times 10^{-4}\) (b) \(2.4 \times 10^{-4}\) (c) \(2.4 \times 10^{4}\) (d) \(1.2 \times 10^{4}\)

An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge \(1.6 \times 10^{-19} \mathrm{C}\) and moving with a velocity of \(10^{6} \mathrm{~m} / \mathrm{s}\) enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)

The deflection in a Galvanometer falls from 50 division to 20 when \(12 \Omega\) shunt is applied. The Galvanometer resistance is (a) \(18 \Omega\) (b) \(36 \Omega\) (c) \(24 \Omega\) (d) \(30 \Omega\)

A bar magnet having a magnetic moment of \(2 \times 10^{4} \mathrm{JT}^{-1}\) is free to rotate in a horizontal plane. A horizontal magnetic field \(\mathrm{B}=6 \times 10^{-4}\) Tesla exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction \(60^{\circ}\) from the field is (a) \(0.6 \mathrm{~J}\) (b) \(12 \mathrm{~J}\) (c) \(6 \mathrm{~J}\) (d) \(2 \mathrm{~J}\)
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