# The partial of field energy with respect to lambda

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the partial of field energy with respect to lambda: This can be rearranged as: Then the force can be expressed: At 8 A, this is to get the energy, we integrate the force over the distance. The antiderivative is: Note this is the energy going into the mechanical system. It is the negative of the energy into the coupling field (since this energy is taken from the coupling field). evaluating from 0.1 to 0.2 m, and negating to get the appropriate energy gives: SP #1 First state the coenergy: The torque is the derivative with respect to theta: With the given currents: The energy transferred from the coupling field to the mechanical system is just this integrated from 0 to pi/2. The question asks for the negative of this (from system to field).

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so Since the field energy and coenergy are the same for this system (linear magnetically), we can just integrate the field energy from one position to the next. This result is obtained by integrating W.c above from 0 to pi/2: The first two terms are really stored energy that doesn't end up getting converted to mechanical energy. The last term is indentical to the energy to energy transferred from the mechanical system. SP #2 Given the following: Going from a-b, x is held constant at 0.03 m and we're given lambda is 8 V-s. therefore, From b-c the current is held constant, but x is changed to 0.02. We can integrate to get field energy, but it is just the same as the coenergy since linear. The force is obtained just by differentiating coenergy (field energy is same in this case).

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This device is acting as a motor. The transition from b-c in the fe-x diagram is the only one with mechanical energy conversion. The change in position is negative (0.03 to 0.02) and the force is negative, resulting in a positive mechanical energy output. This is the definition of a motor.
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