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Chapter 20: Problem 28

A horizontal circular loop carries a current that looks anti-clockwise when viewed from above. It is replaced by an equivalent magnetic dipole \(N \cdot S\). Which of the following is true? (a) The line \(N\)-S should be along a diameter of the loop (b) The line \(N\) - should be perpendicular to the plane of the loop (c) South pole should be below the loop (d) North pole should be below the loop

Short Answer

Expert verified
The south pole should be below the loop.

Step by step solution

01

Understand the Problem

We have a current loop viewed from above as anti-clockwise, which implies the direction of the current flow. We need to replace this with an equivalent magnetic dipole.
02

Clarify Current Loop Direction

Since the loop current is anti-clockwise from above, due to the right-hand rule, the magnetic field will point upwards through the center of the loop.
03

Apply Right-Hand Rule for Magnetic Field

Wrap your right hand around the loop with your fingers pointing in the direction of the current. Your thumb points in the direction of the magnetic field, which in this case is upwards.
04

Determine Equivalent Dipole Orientation

The equivalent magnetic dipole will align with the direction of the magnetic field generated by the loop, which is upwards perpendicular to the plane of the loop.
05

Assign Pole Positions

The north pole of the dipole will be in the direction of the magnetic field (upwards above the loop), and hence, the south pole will be below.
06

Evaluate Options

According to the steps above, option (c), "South pole should be below the loop," is correct.

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Key Concepts

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

Current Loop
In a current loop, electric current continuously flows along a circular path. Imagine it as having little arrows all pointing in the same direction, circling around. When viewed from above, if these arrows flow in an anti-clockwise direction, it's crucial to understand what this means for the loop's magnetic properties.
  • Since the current flows in a circle, it creates a magnetic field similar to a bar magnet.
  • The current loop acts as a magnetic dipole, just like a tiny bar magnet with distinct north and south poles.
  • The concept of an 'equivalent magnetic dipole' refers to the magnetic field the current loop generates, having its own north and south poles.
The transformation of a current loop into a magnetic dipole involves identifying the direction of the magnetic field it creates, which relies on the associated magnetic rules.
Right-Hand Rule
The right-hand rule is a helpful tool for determining the direction of the magnetic field created by a current. It's a simple and intuitive way to connect the flow of current and the resulting magnetic field.

Follow these steps:
  • With the current flowing anti-clockwise from above, imagine wrapping your right hand around the loop.
  • Your fingers should curve in the direction of the current, as they do around the wire or loop.
  • Your thumb, extended like you're giving a thumbs-up, will point in the direction of the magnetic field.
In this exercise, this means your thumb points upwards when the loop is held horizontally. Consequently, the magnetic field generated by this current loop moves upwards, perpendicular to the loop's plane. This orientation helps to establish the north and south poles of the equivalent dipole.
Magnetic Field Direction
Understanding the magnetic field direction is key in determining where the north and south poles lie in a magnetic dipole. The direction of the magnetic field is the direction that the north end of a compass needle would point if placed near the loop.
  • For an anti-clockwise current viewed from above, the magnetic field points upwards.
  • This upward direction is perpendicular to the plane of the loop.
  • The direction aligns with the right-hand rule prediction.
Here, the upward-pointing magnetic field signifies that the north pole of the equivalent magnetic dipole is above the plane of the loop, with the south pole being below the plane. This explains why the correct choice in our original exercise is that the south pole is below the loop, meaning option (c) is correct.

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Most popular questions from this chapter

At a certain location in Africa, a compass points \(12^{\circ}\) west of the geographie north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points \(60^{\circ}\) above the horizontal. The horizontal component of the earth's field is measured to be \(0.16\) G. Specify the direction and magnitude of the earth's field at the location. (a) \(32^{\circ}\) west of geographical meridian and \(3.2 \times 10^{-4} \mathrm{~T}\) (b) \(12^{\circ}\) west of geographical meridian and \(0.32 \times 10^{-4} \mathrm{~T}\) (c) \(12^{2}\) east of geographical meridian and \(0.32 \times 10^{-4} \mathrm{~T}\) (d) 32 ' cast of geographical meridian and \(3.2 \times 10^{-4} \mathrm{~T}\)

A thin bar magnet of length \(2 L\) is bent at the mid-point so that the angle between them is \(60^{\circ}\). The \mathrm{\\{} n e w ~ l e n g t h ~ o f ~ t h e ~ m a g n e t ~ i s ~ [BVP Engg. 2008] (a) \(\sqrt{2} L\) (b) \(\sqrt{3} L\) (c) \(2 \bar{L}\) (d) \(L\)

A bar magnet having a magnetic moment of \(2 \times 10^{4} \mathrm{JT}^{-1}\) is free to rotate in a horizontal plane. \(\mathrm{A}\) horizontal magnetic field \(B=6 \times 10^{-4} \mathrm{~T}\) 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 [Kerala CET 2009] (a) \(2 \mathrm{~J}\) (b) \(0.6\rfloor\) (c) \(12 \mathrm{~J}\) (d) \(6 \underline{1}\)

At the magnetic north pole of the earth, the value of the horizontal component of earth's magnetic field and angle of dip are respectively (a) zero, maximum (b) maximum, minimum (c) maximum, maximum (d) minimum, minimum

In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth's magnetic field is \(2 \mathrm{~s}\). When a magnet is brought near and parallel to it, the time period reduces to 18 . The ratio \(F / H\) of the fields, \(F\) due to magnet and \(H\), the horizontal component will be (a) \(\sqrt{3}\) (b) \(\frac{1}{\sqrt{3}}\) (c) \(\frac{1}{3}\) (d) 3
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