Maximum potential of a sliding box down a hill

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SUMMARY

The discussion focuses on calculating the maximum potential induced in a wire loop as a 150 kg box slides down a 4.35 m hill at a 30.0° angle, encountering a magnetic field produced by a solenoid with 5000 loops and a current of 1.40 A. The coefficient of kinetic friction is 0.270. The magnetic field strength is calculated to be 0.00704 T. The induced current's direction and maximum amount through the loop, which has an internal resistance of 750 Ω, are also key points of inquiry.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with Newton's laws of motion
  • Knowledge of friction coefficients and their impact on motion
  • Basic concepts of magnetic fields and solenoids
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  • Calculate the speed of the box upon reaching the horizontal plane
  • Explore Faraday's Law of electromagnetic induction
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1. Maximum potential of a sliding box down a hill
1. A 150 kg box (with dimensions of 1.30 m on a side) sits at the top of a hill 4.35 m long angled at 30.0° to the horizontal. The box has 250 loops of wire wrapped around it. The loops lie in the plane of this paper. The box is allowed to slide down the hill, which has a coefficient of kinetic friction of 0.270. Once the box reaches the bottom of the hill, it slides along a horizontal frictionless plane, where it encounters a magnetic field produced by a 1.25 m long solenoid with 5000 loops and a current of 1.40 A running through it. The magnetic field points out of the page.
a. What is the maximum potential produced in the wires wrapped around the box as it passes through the magnetic field? (The size of the field is large compared to the size of the box)
b. In what direction does the induced current flow?
c. If the wires have an internal resistance of 750 Ω, what is the maximum amount of induced current that flows through the loop?


2. i don't understand what to do after this point
3.
150 kg
l=1.3m
A=1.69m^2
n=250 loops
u=0.270
B= out of the page
l=1.25m
n=5000 turns
I=1.4 A

B(horz)= (u(o) NI) / l
B(horz)= [(4*pi*10^-7)(5000)(1.4A)]/1.25M
B(horz)=0.00704 T
 
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