# 22 the coil moves in a uniform magnetic field b with

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A rectangular coil with resistance R has N turns, each of length l and width w as shown in Figure P23.22. The coil moves in a uniform magnetic field B with constant velocity v. What are the magnitude and direction of the total magnetic force on the coild as it (a) enters, (b) moves within, and (c) leaves the magnetic field. (a) As in Problem 13, d Φ B / dt = wBv , so the induced current is I = ε R = - d Φ B / dt R = - wvBN R (22) Where the - sign indicates it is counterclockwise (against the changing flux direction). And the force on the leading wires is F = I l × B = - I · Nw · B ˆ i = - v ( wBN ) 2 R ˆ i (23) (b) Once the coil is inside the magnetic field, the flux becomes constant, so there is no induced emf driving a current, and thus no net force on the coil. (c) The situation here is the inverse of that in (a) , so the induced emf is clockwise, but the current through the portion of loop in the magnetic field is still up , so the force is unchanged. F = - v ( wBN ) 2 R ˆ i (24) Problem 53. A particle with a mass of m = 2 . 00 · 10 - 16 kg and a charge of q = 30 . 0 nC starts from rest, is accelerated by a strong electric field, and is fired from a small source inside a region of uniform constant magnetic field B = 0 . 600 T. The velocity of the particle is perpendicular to the field. The circular orbit of the particle encloses a magnetic flux of Φ B = 15 . 0 μWb. (a) Calculate the speed of the particle. (b) Calculate the potential difference through which the particle accelerated inside the source. (a) For particles circling in a uniform, perpendicular magnetic field, F c = m v 2 r = qvB (25) mv = qrB (26) Letting τ be the period, from Δ x = v Δ t we have τ = 2 π r v = 2 π rm qrB = 2 π m qB = 69 . 8 ns (27) The inverse of our cyclotron frequency from Recitation 7.

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• Spring '08
• TUSSEY
• Physics, Magnetic Field, ε

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