Magnetic Field and range of potential energy

AI Thread Summary
The discussion focuses on calculating the potential energy of a circular wire in a uniform magnetic field. Given a radius of 0.02m, a magnetic field strength of 1*10^-4 T, and a current of 5A, the torque is expressed as torque = BIA. The potential energy is derived from the relationship between torque and potential energy, specifically through the equation for torque as a function of the angle between the magnetic field and the wire. The solution indicates that potential energy can be determined by integrating the torque with respect to the angle. Understanding this relationship is crucial for solving the problem effectively.
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Homework Statement


A wire is formed into a circle having a radius of .02m and placed in a uniform magnetic field of 1*10^-4. A current of 5A passes through the wire. What is the range of potential energy the wire possesses for different orientations?


Homework Equations


torque=BIA



The Attempt at a Solution



The answer is 2BIA. However, I am not sure how to relate torque to potential energy in this scenario.
 
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torque is actually equal to BIAsin(theta), where theta is the angle between the field and the area. Potential energy is just the integral of -Td(theta), so find the function for potential energy and go from there.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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