ps07soln - Physics 217 Homework 7 Solutions Ruxandra...

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Unformatted text preview: Physics 217 - Homework 7 Solutions Ruxandra Bondarescu April 4, 2005 Problem 1. Purcell 6.14 We set a coordinate system so the point P under consideration is in the z- x plane with coordinates ( x o , ,z o ). Consider a current loop at an angle φ from the x axis. One small section of the loop has coordinates ( r cos φ,r sin φ,z ) and r = ( x o- r cos φ,- r sin φ,z o- z ) (1) The direction of this little current is arbitrary within the plane so the current direction vector can be written I = I r ˆ r + I z ˆ z = ( I r cos φ,I r sin φ,I z ) (2) Then the contribution to the magnetic field will be in the direction I × r = [sin φ ( I r ( z o- z ) + rI z )] ˆ x + [ I z ( x o- r cos φ )- I r cos φ ( z o- z )] ˆ y- I r x o sin φ ˆ z (3) However, for each section of current, there is a similar section that is identical except that φ → - φ . We see that the x and z components change sign. Therefore, these components cancel out and the net field at the point P is in the y direction. For our coordinate system, this in the circumferential direction. Once we have this information, it is easy to use Ampere’s law. integraldisplay B · dlscript = 2 πrB = 4 π c NI (4) B = 2 NI cr , a < r < b (5) If we are outside the solenoid, the enclosed current will be zero and so the field is zero. Problem 2. Problem 7.2 The magnetic field will point only in the z direction B z = B cos ( θ ) since the wire is parallel to the y-axis and the velocity of the loop points in the...
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ps07soln - Physics 217 Homework 7 Solutions Ruxandra...

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