In this problem, you will explore a simplified model of magnetic braking. A roll

Question

In this problem, you will explore a simplified model of magnetic braking. A roller coaster car of total mass 10000 kg has a coil of N = 550 turns attached to its undercarriage. This coil is rectangularly shaped, with dimensions of L = 4.7 m along the direction of motion and W = 1 m wide, and with total resistance 500 ohms

When the car is to be stopped from an initial speed vi = 20 m/s while travelling in the +x direction, a non-uniform magnetic field pointing upward is suddenly turned on in the region of space where the coil is. It varies in strength along the direction of motion, according to the equation: B(x) = b1 · x + b2 where constants: b1 = 0.46 T/m and b2 = 0.1 T. I.)

II.) At the moment the field is turned on:

a.) What is the induced current? Hint #1: Any linear function has an average value halfway between the endpoints. This can be used to greatly expedite your flux calculation. Hint #2: Use the chain rule to switch to a better variable to take the derivative of.

initial current Induced =??? A

b.) Though this magnetic field is a function of x, it turns out to be unimportant exactly where the coil is relative to the origin (x=0), as long as it is entirely within the prescribed field. This is because the difference between the magnetic field from the front and the back of the coil is always

= ?????T

c.) What is the size of the deceleration of the car initially?

initial a =??????? m/s^2

III.) Later on,

d.) Show that the acceleration of the car as a function of time takes on the form: a(t) = - C*v(t)

Find the constant "C":

C = ?????s-1

e.) Use this function to find the time it will take to slow down the car to 5% of its initial velocity:

t=????? s

Explanation / Answer

we have

mass= 10000 kg

N=550

R=500 ohm

L=4.7 m W= 1m

Vi= 20 m/s

B=0.46 x A +0.1

we have '

1).voltage induced = d(phi)/dt

where phi is flux

now flux= N*B*A

B at midpoint is i.e x= 4.7/2 is 0.46*4.7/2 +.1=1.81 T

so induced voltage is NA*dB/dt

now dB/dt= 0.46*dx/dt

= 0.46v

so induced emf= 550*4.7*1*0.46*20

23782 V

so induced current= E/R

= 23782/500

= 47.564 A

B). difference is 0.46*4.7=2.162 N

c)F=NILB

= -550*47.564*1*2.162

= -56558.35 N

so a= F/m

=-5.6556 m/s^2

d) a(t)=

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