/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A single-turn square loop of wir... [FREE SOLUTION] | 91Ó°ÊÓ

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A single-turn square loop of wire \(2.00 \mathrm{~cm}\) on a side carries a counterclockwise current of \(0.200 \mathrm{~A}\). The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30 turns per centimeter and carries a counterclockwise current of \(15.0 \mathrm{~A}\). Find the force on each side of the loop and the torque acting on the loop.

Short Answer

Expert verified
The force on each side of the loop is \( 0.000226 N \). The torque acting on the loop is 0 N.

Step by step solution

01

Calculate the magnetic field inside the solenoid

To calculate the magnetic field inside the solenoid, we need to use the formula \[ B = \mu_0ni \]. Here \( n = 30 /cm \) or \( 3000/m \), \( i = 15 A \), and \( \mu_0 = 4\pi \times 10^{-7} Tm/A \). Substituting these values, we get: \[ B = (4\pi \times 10^{-7} Tm/A) \times 3000/m \times 15 A = 0.0565 T \]
02

Calculate the force on each side of the loop

We can now use the magnetic field inside the solenoid to find the force on each side of the loop. The force can be calculated using the formula: \(F = I \times (L \times B)\), where \(I\) is the current in the wire, \(L\) is the length of side, and \(B\) is the magnetic field. Since \(L\) and \(B\) are perpendicular, the magnitude will be \( F = I \times L \times B \) Substituting \(I = 0.2 A\), \(L = 0.02 m\), and \(B = 0.0565 T\), we obtain: \( F = 0.2 A \times 0.02 m \times 0.0565 T=0.000226 N\)
03

Calculate the torque on the loop

The torque on the loop can be calculated using the formula: \[ T = r \times F \] where \( r \) is the distance from the axis to the point of application of force and \( F \) is the magnitude of the force. However, in this case, the torque will be zero because the loop is symmetrically placed in the magnetic field, and the forces on opposite sides of the loop are equal in magnitude and opposite in direction, thus canceling out any torque.

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Key Concepts

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

Magnetic Field Inside a Solenoid
Understanding the magnetic field inside a solenoid is fundamental when dealing with electromagnetism. A solenoid is essentially a long coil of wire, with many turns, through which an electric current passes. The magnetic field created inside a solenoid is very uniform, and its strength can be adjusted by changing either the current flowing through the wire or the number of turns per unit length of the solenoid.

Using Ampère's Law, we can derive the magnetic field inside an ideal solenoid as \[ B = \boldsymbol{\frac{{\boldsymbol{\bar{\boldsymbol{i}} \times \boldsymbol{\bar{i}}}}}{{\boldsymbol{\bar{l}}}}} \brush\examplefont\brush\], where \( \boldsymbol{\bar{i}} \brush\hangulfont\brush\homer\) is the number of turns per unit length (turns per meter), \( \boldsymbol{\bar{i}} \brush\blackboardfont\brush\dickens\) is the current, and \( \boldsymbol{\bar{\mu_0}} \brush\shadowfont\brush\da\) is the permeability of free space. This formula shows how the magnetic field can become very strong if either the current or the number of turns is sufficiently large.

To grasp this concept fully, imagine wrapping a wire around a pencil. The wire turns represent the coils of the solenoid, and when you pass a current through the wire, a magnetic field appears inside the pencil, analogous to the field inside a solenoid. If you wrap the wire more tightly or increase the current, the magnetic field becomes stronger, just as it does in our solenoid example. This principle is widely used in electromagnets, transformers, and even in medical imaging techniques such as MRI.
Lorentz Force
The Lorentz force is a crucial concept when discussing the interaction between magnetic fields and current-carrying conductors. The force is named after the Dutch physicist Hendrik Lorentz. It is the combined electric and magnetic force on a point charge due to electromagnetic fields. However, when dealing with a current-carrying wire such as the loop in our exercise, we're primarily concerned with the magnetic component of this force.

The force on a small segment of wire carrying current in a magnetic field is given by the equation \[ F = I \times (L \times B) \brush\cjkfont\brush\feedin\], where \( I \brush\licenseplatefont\brush\eb\) is the current, \( L \brush\artdecofont\brush\hollyhock\) represents the length of the current segment, and \( B \brush\artnouveaufont\brush\metal\) is the magnetic field. The cross product suggests that the force's direction is perpendicular to both the direction of the current and the magnetic field, following the right-hand rule. For our square loop, this means the force acts inward or outward of the plane defined by the loop, trying to expand or compress it, depending on the current's direction.

An intuitive way to visualize the Lorentz force is to imagine you're pushing a toy car (representing the charge) along a track (the wire). When you introduce a magnet to one side of the track, it will either pull the car towards it or push it away, depending on the poles of the magnet and the direction you were moving the car. This effect is analogous to the Lorentz force acting on the current within the wire.
Torque on a Current Loop
When a current loop is placed in a magnetic field, each segment of the loop experiences a magnetic force. These forces can cause the loop to rotate, creating a torque. The torque (\( T \)) on a current loop is given by the equation \[ T = r \times F \brush\grungefont\brush\fantasy\], where \( r \brush\westernfont\brush\motel\) is the position vector from the center of the loop to the point where the force is applied, and \( F \brush\easternfont\brush\race\) is the force due to the magnetic field. The direction of this torque is determined by the right-hand rule and will be such that it tends to rotate the plane of the loop to align with the magnetic field.

In our particular exercise, the loop is symmetrically placed within the magnetic field of the solenoid, and hence, the forces on opposite sides are equal and opposite. As a result, they produce no net torque; the loop would not rotate despite the magnetic forces. This is a specific scenario resulting from the symmetrical placement of the loop. However, if the loop were positioned differently, these forces could cause it to rotate, revealing the fundamental principle behind electric motors.

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Most popular questions from this chapter

A current of \(17.0 \mathrm{~mA}\) is maintained in a single circular loop with a circumference of \(2.00 \mathrm{~m}\). A magnetic field of \(0.800 \mathrm{~T}\) is directed parallel to the plane of the loop. What is the magnitude of the torque exerted by the magnetic field on the loop?

It is desired to construct a solenoid that will have a resistance of \(5.00 \Omega\) at \(20^{\circ} \mathrm{C}\) ) and produce a magnetic field of \(4.00 \times 10^{-2} \mathrm{~T}\) at its center when it carries a current of \(4.00 \mathrm{~A} .\) The solenoid is to be constructed from copper wire having a diameter of \(0.500 \mathrm{~mm}\). If the radius of the solenoid is to be \(1.00 \mathrm{~cm}\), determine (a) the number of turns of wire needed and (b) the length the solenoid should have.

A wire \(2.80 \mathrm{~m}\) in length carries a current of \(5.00 \mathrm{~A}\) in a region where a uniform magnetic field has a magnitude of \(0.390 \mathrm{~T}\). Calculate the magnitude of the magnetic force on the wire, assuming the angle between the magnetic field and the current is (a) \(60.0^{\circ}\), (b) \(90.0^{\circ}\), and (c) \(120^{\circ}\).

A \(0.200-\mathrm{kg}\) metal rod carrying a current of \(10.0 \mathrm{~A}\) glides on two horizontal rails \(0.500 \mathrm{~m}\) apart. What vertical mag netic field is required to keep the rod moving at a constant speed if the coefficient of kinetic friction between the rod and rails is \(0.100\) ?

Using an electromagnetic flowmeter (Fig. P19.69), a heart surgeon monitors the flow rate of blood through an artery. Electrodes \(A\) and \(B\) make contact with the outer surface of the blood vessel, which has interior diameter \(3.00 \mathrm{~mm} .\) (a) For a magnetic field magnitude of \(0.0400 \mathrm{~T}\), a potential difference of \(160 \mu \mathrm{V}\) appears between the electrodes. Calculate the speed of the blood. (b) Verify that electrode \(A\) is positive, as shown. Does the sign of the \(\mathrm{emf}\) depend on whether the mobile ions in the blood are predominantly positively or negatively charged? Explain.

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