QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
A circular coil 17.0 cm in radius and composed of 145 tightly wound turns carries a current of 2.30 A in the counterclockwise direction, where the plane of the coil makes an angle of 15.0° with the y axis (see figure below). The coil is free to rotate about the z axis and is placed in a region with a uniform magnetic field given by \vec{B} = 1.25 \, \vec{f} \, t .
(a) What is the magnitude of the magnetic torque on the coil?
\mu = 1 \, \text{N} \times \text{m}
(b) In what direction will the coil rotate?
- clockwise as seen from the +z axis
- counterclockwise as seen from the +z axis
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Answer
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Step 1: Find the magnetic moment of the coil.
μ = NAI = 145 × 0.0909 m^2 × 2.30 A = 29.85 A·m^2
The magnetic moment (μ) of a coil can be calculated using the formula: where N is the number of turns, A is the area of the coil, and I is the current. First, we need to find the area (A) of the coil. The radius of the coil is given as 17.0 cm, so its diameter is 34.0 cm. The area of the circle is given by: where r is the radius. Substituting the given radius into this formula, we get: Now, we can calculate the magnetic moment using the given current (I = 2.30 A) and the number of turns (N = 145):
Step 2: Calculate the magnetic torque on the coil.
τ = μB = 29.85 A·m^2 × 1.25 T = 37.31 N·m
The magnetic torque (τ) on a coil in a magnetic field can be calculated using the formula: where μ is the magnetic moment and B is the magnetic field. The magnetic field B is given as: To calculate the torque, we first need to find the component of the magnetic field along the axis of rotation (z-axis). Since the magnetic field is in the x-direction and the coil's axis makes a 15° angle with the y-axis, the component of the magnetic field along the z-axis is zero. However, the problem asks for the magnitude of the torque, which is a scalar value. The magnitude of the magnetic moment (μ) is already given, and the magnitude of the magnetic field (B) is also given. Therefore, the magnitude of the torque is simply the product of these two values:
Final Answer
(a) The magnitude of the magnetic torque on the coil is 37.31 N·m. Step 3: Determine the direction of rotation. To determine the direction of rotation, we need to find the direction of the torque. The torque acts in the direction perpendicular to both the magnetic moment and the magnetic field. In this case, the magnetic moment is along the z-axis (out of the page), and the magnetic field is in the x-direction. Therefore, the torque is in the y-direction. Since the coil is free to rotate about the z-axis, it will rotate in the direction that aligns its magnetic moment with the magnetic field. In this case, the magnetic field is increasing in the positive x-direction. Therefore, the coil will rotate in the counterclockwise direction as seen from the +z axis to align its magnetic moment with the magnetic field. (b) The coil will rotate counterclockwise as seen from the +z axis.
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