Homework_Ch_23_solutions

# Homework_Ch_23_solutions - P202 Spring 2006 Ch 23 Magnetic...

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P202 Spring 2006 Ch 23: Magnetic Flux and Faraday’s Law Homework due at 11:59 pm Feb. 19, 2006 Error allowance on all numerical answers is 3%. 1. The magnetic field produced by an MRI solenoid L=[2.0 – 2.5] m long and D=[1.0 – 1.5] m in diameter is B=[1.7 – 2.1] T. (a) What is the magnetic flux through the solenoid? _____________ Wb (b) What is the magnetic energy stored in the device? ____________ J Solution: (a) Find the magnetic field inside the solenoid using eqn (22-12) of Walker. Use this field to find the flux through the solenoid using eqn (23-1) of Walker, noting that the magnetic field of the solenoid is along the solenoid axis so cos θ = 1. (b) The magnetic energy density is given by eqn (23-20) of Walker. We already calculated the magnetic field within the solenoid in part (a) so all we need to do is to plug this field into eqn (23-20) to get the energy density. The total energy is given by the energy density times the volume of the solenoid which we can easily find from its diameter and length Solution: (a) = 0.7854B D 2 (b) = 3.125 D 2 L B 2 10 5 2. A magnetic field increases linearly from 0 to B=[0.1 – 0.2] T in t=[1.0 – 2.0] s. How many turns of wire are needed in a circular coil D=[8.0 – 12.0] cm in diameter to produce an induced EMF of v=[5.0 – 6.0] V. Assume the plane of the coil is perpendicular to the field._______________ Solution: We are told that the magnetic field changes linearly by a certain amount (B) in a certain time (t). Thus we know the rate of change of B. Knowing this, we can calculate the rate of change of magnetic flux using eqn (23-1) of Walker. We are told that the coil is perpendicular to the magnetic field so cos = 1 in this eqn. The area A in eqn (23-1) is just the area of the coil. Having calculated the rate of change of flux, we can plug this into eqn (23-4) of Walker to obtain an expression for the EMF induced in the coil. The unknown in this eqn is the number of turns for the coil, for which we can solve. Solution: =12732.0 (vt/BD 2 ) 3. An airplane with a wing span of L =[30.0 – 40.0] m flies at a speed of v=[600 – 850] km/h in a north westerly direction. (a) If the vertical component of the Earth’s magnetic field is 5 x 10 -6 T, what is the induced EMF between the wing tips.__________ mV

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(b) If the plane turns to head due north does the induced EMF increase [radio button], stay the same [radio button] or increase [radio button] Solution: (a) The trick here is to note that the horizontal component of the Earth’s magnetic field is in the same plane at the velocity vector of the aircraft and the vector across the wing span of the aircraft. Thus, this component of the Earth’s field produces no induced EMF between the wing tips (although it does produce
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Homework_Ch_23_solutions - P202 Spring 2006 Ch 23 Magnetic...

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