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Chapter 26: Problem 55

A simple electric motor consists of a 100 -turn coil \(3.0 \mathrm{cm}\) in diameter, mounted between the poles of a magnet that produces a \(0.12-\mathrm{T}\) field. When a 5.0 -A current flows in the coil, what are (a) its magnetic dipole moment and (b) the motor's maximum torque?

Short Answer

Expert verified
The magnetic moment of the coil is given by \(m = 100 × 5.0 A × π(0.015^2) A.m^2\). The maximum torque on the coil is \(\tau_{max} = 100 × 5.0 A × π(0.015^2) × 0.12 T\).

Step by step solution

01

Compute the area of the coil

The area \(A\) of the circular coil can be calculated using the formula \(A = π(r^2)\), where \(r\) is the radius of the coil. In this case, the diameter of the coil is given as 3.0 cm, therefore the radius \(r\) is 1.5 cm or 0.015 m. Substituting these values into the formula, yields \(A = π(0.015^2)\) m^2.
02

Calculate the magnetic dipole moment

The magnetic dipole moment \(m\) can be calculated using the formula \(m = nIA\), where \(n\) = number of turns, \(I\) = current and \(A\) = area. In this case, there are 100 turns, the current is 5.0 A and we have calculated the area in the last step. Substituting these values into the formula gives \(m = 100 × 5.0 A × π(0.015^2)\) A.m^2.
03

Calculate the maximum torque

The maximum torque on the coil, \(\tau_{max}\), can be calculated using the formula \(\tau_{max} = mB\), where \(m\) is the magnetic dipole moment and \(B\) is the magnetic field strength. The torque is maximum when the angle between the magnetic moment and the magnetic field, \(θ\), is 90 degrees and sin(90)=1. Thus, substituting the values in the formula gives \(\tau_{max} = 100 × 5.0 A × π(0.015^2) × 0.12 T\).

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Most popular questions from this chapter

An electric motor contains a 250 -turn circular coil \(6.2 \mathrm{cm}\) in diameter. If it develops a maximum torque of \(1.2 \mathrm{N} \cdot \mathrm{m}\) at a current of \(3.3 \mathrm{A},\) what's the magnetic field strength?

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