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CAF123
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Homework Statement
A coil is formed by wrapping 80 cm of thin wire round a non-conducting square of size 2 × 2 cm which is placed in a horizontal magnetic field of strength 20mT.
A)Sketch the orientation of the coil that will result in the maximum torque acting, and the value of this torque when the coil is carrying a current of 200mA.
B) Explain how the above system can be modified to produce a simple electric motor where the coil rotates in the magnetic field.
The Attempt at a Solution
A)We have derived in class that the torque on a coil suspended in a B field is ##\tau = AIB\sin \theta##. So this will be maximal when ##\theta = \pi/2## which means ##\tau = (4 \times 10^{-4} m^2 \cdot (200 \times 10^{-3} A) \cdot (20 \times 10^{-3})T \approx 1.6 \times 10^{-6}Nm##
Is this correct? I haven't used the length of the wire here but I guess I assumed it covered the whole sqaure. If it is 80cm, then each full turn will give 4cm of the wire used which means to wrap the whole wire around the sqaure I need 20 turns. However, I don't know the thickness of the wire so I am unable to tell how much of the 2x2 square those 20 turns take up. Is that 80cm a distractor?
B) I get the general idea about having to switch the direction of current every 180 degrees so as to reverse the magnetic force acting on the coil. But I have a question. Say the B field is horizontal to the left. Put the square parallel to the field such that a current runs clockwise. Then the magnetic force tends to tilt the sqaure upwards until it is perpendicular to the B field. The torque is then zero and the square will fall back down. How then is it possible to get the square to exceed 90 + degrees? I thought it could simply be inertia which makes it just exceed the 90 mark, but then if this is the case then since the current in the square is now flipped, this tends to push the sqaure back towards the 90 mark. So how is it even possible to get the sqaure to 180 degrees? (I hope this makes sense, if not I will attach a sketch).
Many thanks.