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Chapter 19: Problem 57

A certain fixed length \(L\) of wire carries a current \(I\) (a) Show that if the wire is formed into a square coil, then the maximum torque in a given magnetic field \(B\) is developed when the coil has just one turn. (b) Show that the magnitude of this torque is \(\tau=\frac{1}{16} L^{2} I B\)

Short Answer

Expert verified
Answer: The magnitude of the maximum torque produced by a square coil containing a single turn is given by the expression \(\tau_\text{max} = \frac{1}{16} L^{2} I B\) where \(L\) is the total length of the wire, \(I\) is the current flowing through the wire, and \(B\) is the magnetic field strength.

Step by step solution

01

Determine the total turns and the length of each side of the square coil

Let's assume we have \(n\) turns of the wire in the square coil, and each side of the square is \(a\). To find the maximum torque, we need to determine the relationship between \(a\), \(n\) and the total length of the wire, \(L\). Since there are \(n\) turns, the total length of the wire can be written as the perimeter of the square multiplied by the number of turns: $$ L = 4an $$
02

Calculate the magnetic moment of the square coil

The magnetic moment \(\mu\) of the square coil is given by the product of the current \(I\), the number of turns, and the area of the square loop: $$ \mu = nIA $$ The area of the square loop can be expressed as: $$ A = a^2 $$ Substituting this into the magnetic moment, we get: $$ \mu = nIa^2 $$
03

Express the torque in terms of L, n and B

The torque \(\tau\) on a loop in a magnetic field is given by: $$ \tau = \mu B\sin{\theta} $$ In this case, we will consider the angle \(\theta\), at which the torque is maximum: $$ \sin{\theta_\text{max}} = 1 $$ Thus, the maximum torque can be written as: $$ \tau_\text{max} = \mu B $$ By substituting the expression for the magnetic moment from step 2, we have: $$ \tau_\text{max} = nIa^2 B $$ Now, let's express \(a^2\) in terms of \(L\) and \(n\) using the equation from step 1: $$ a^2 = \frac{L^2}{16n^2} $$ Substitute this expression into the torque equation: $$ \tau_\text{max} = \frac{1}{16}nIBL^2 $$
04

Find the value of n that maximizes the torque

To find the value of \(n\) that maximizes the torque, we will differentiate the torque expression with respect to \(n\) and set it equal to zero: $$ \frac{d\tau_\text{max}}{dn} = 0 $$ $$ \frac{d}{dn}\left(\frac{1}{16}nIBL^2\right) = 0 $$ $$ \frac{1}{16}IBL^2 = 0 $$ Since \(IBL^2\) is a constant and cannot be zero, the maximum torque occurs when \(n=1\). Therefore, the maximum torque occurs when the wire is formed into a square coil with just one turn.
05

Calculate the magnitude of the maximum torque

Now that we know the maximum torque occurs when there is only one turn in the square coil, we can substitute \(n=1\) in the torque expression: $$ \tau_\text{max} = \frac{1}{16}(1)IBL^2 = \frac{1}{16} L^2 I B $$ Thus, the magnitude of the maximum torque is \(\boxed{\tau_\text{max} = \frac{1}{16} L^{2} I B}\).

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