Calculating Current for Electromagnet Lift

AI Thread Summary
To calculate the current needed to lift a Cadillac weighing 4200 N using an electromagnet, the relationship between force, magnetic field strength, and current must be understood. The initial current of 200 A is sufficient to lift a 360 kg car, but the required force for the Cadillac is significantly higher. The discussion emphasizes that the magnetic field strength is proportional to the current, suggesting that the current must be increased proportionally to the force needed. Participants are encouraged to clarify their equations and concepts, particularly the distinction between electric charge and current. Ultimately, the solution involves applying the correct principles of electromagnetism to determine the necessary current for the lift.
29 clicks
Messages
15
Reaction score
0
I have this question that I have been completely stuck on with the equations that i have and i was wondering if someone could help me with this question but what equation should i use first to get myself going.

Someone drives an electromagnet at a junk yard. Normally the magnet uses 200 A to lift a car of 360 kg, but now a force of 4200 N is required to lift a big old cadillac. What current would be required to make this lift?
 
Physics news on Phys.org
Perhaps you could start by showing us which equations you have to work with. Please share with us some of your thoughts and the work you have done. We can then guide you to a solution.
 
This is what i have so far.

I = 200 A
Car Mass = 360 kg
F = 4200 N
Q= ?

I'm looking for the current required to lift the car. The only equation i know that can find is Q= I delta t and Q=Ne but i don't see how any of those equations would help. If you see something i can't then show me what i should do.
 
First of all, Q refers to "Quantity of Electric Charge", not "Indicated Current".

Do you know what causes a magnetic field in the electro-magnet?
(hint: it is NOT electric charge ...)
You're probably supposed to presume that the strength of the megnet
is proportional to the quantity that determines the magnetic field.
 
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}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top