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au10_312ps1_sol

# au10_312ps1_sol - ECE312 Autumn 2010 Problem Set 1...

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Unformatted text preview: ECE312 Autumn 2010: Problem Set 1 Solutions Problem 1 : B = 5 μ ˆ y T, u = 3ˆ x + 4ˆ z m/s, m e = 9 . 1 × 10 − 31 kg, and q e =- 1 . 6 × 10 − 19 C. (a) The force on the electron: F = q u × B = (- 1 . 6 × 10 − 19 )(3ˆ x + 4ˆ z ) × 5 μ ˆ y = 5(- 1 . 6 × 10 − 19 )(4 π × 10 − 7 )(3ˆ z- 4ˆ x ) = (5 . 03 × 10 − 24 )(0 . 8ˆ x- . 6ˆ z ) N (b) The radius of the circular orbit: vextendsingle vextendsingle F vextendsingle vextendsingle = 5 . 03 × 10 − 24 = m | u | 2 r ⇒ r = (9 . 1 × 10 − 31 )(5) 2 (5 . 03 × 10 − 24 ) = 4 . 52 μ m (c) Given a circular orbit, the distance traveled in one orbit is 2 πr . Since the velocity | u | is 5 m/s, the time required to complete one orbit is 2 πr | u | = 5 . 68 μ s ⇒ f = (5 . 68 × 10 − 6 ) − 1 = 175 . 94 kHz This is called the “gyrofrequency” of the electron. Electrons circulating in these sorts of orbits are often used in devices for studying particle physics. Problem 2 : A coaxial cable with J = ˆ z (1+ r 2 ) flowing through a solid inner cylinder of radius 2 cm. The current returns on the outer shell of radius 5 cm. (a) Note this is a non-uniform current density, i.e. the current is less in the middle of the wire than on the outside following the function (1+ r 2 ). To use Ampere’s Law to find fields, we need a lot of symmetry. Here we have a cylindrical structure, so we choose a cylindrical coordinate system for this problem. In general the field would look like H ( r, φ, z ) = ˆ rH r ( r, φ, z ) + ˆ φH φ ( r, φ, z...
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au10_312ps1_sol - ECE312 Autumn 2010 Problem Set 1...

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