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Chapter 26: Problem 35

A single-turn square wire loop \(5.0 \mathrm{cm}\) on a side carries a \(450-\mathrm{mA}\) current. (a) What's the loop's magnetic dipole moment? (b) If the loop is in a uniform 1.4 -T magnetic field with its dipole moment vector at \(40^{\circ}\) to the field, what's the magnitude of the torque it experiences?

Short Answer

Expert verified
The loop's magnetic dipole moment is \(0.001125\, A\cdot m^{2}\) and it experiences a torque of \(0.0006055 N\cdot m\) in the given magnetic field.

Step by step solution

01

Calculate the Magnetic Dipole Moment

First, calculate the area of the square loop. The formula to find the area of a square is simply \(A = s^{2}\), where \(s\) is the length of each side. In this case, each side length is \(5.0\, cm = 0.05\, m\). So, the area can be calculated as \(A = 0.05\, m * 0.05\, m = 0.0025\, m^{2}\). Then, the magnetic dipole moment is calculated as the product of the current and the area, following the formula \(\mu = I * A\). Substituting the given values, \(\mu = 0.45\, A * 0.0025\, m^{2} = 0.001125\, A\cdot m^{2}\).
02

Calculate the Torque

Now that the magnetic dipole moment is known, the torque experienced by the loop can be calculated. The torque on a magnetic dipole is given by the formula \(\tau = \mu B \sin \theta\), where \(B\) is the magnetic field strength, \(\mu\) is the magnetic dipole moment, and \(\theta\) is the angle between the dipole moment vector and the magnetic field. Substituting the given and calculated values: \(\tau = 0.001125\, A\cdot m^{2} * 1.4\, T * \sin(40^{\circ}) = 0.0006055 N\cdot m \).

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