hw6.11 - point where the particle exits the magnetic field...

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HW 6.11 Given: F i n d : s m v / 10 5 . 2 6 0 ? 0 v m d 2 . 1 1 ? R / T B 01 . 0 ? m q 11 ? 2 d A charged particle with initial velocity v 0 = 2.5 × 10 6 m/s in the positive x -direction enters a region of depth d 1 = 1.2 m that has a uniform magnetic field B = 0.01 T in the positive z -direction (out the page). The magnetic field is zero elsewhere. The particle leaves the magnetic field region with a velocity vector at an angle = 11 of x -axis. o with respect to the (a) What is the magnitude v 0 ' of the particle's velocity when it exits the magnetic field region? (b) What is the radius of curvature R of the particle's trajectory when in the region of the magnetic field? (c) Calculate the ratio q / m of the charge to the mass of the particle. Be sure to include the correct algebraic sign in your answer. (d) Calculate the displacement d 2 in the y -direction of the particle from its original trajectory at the
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Unformatted text preview: point where the particle exits the magnetic field region. Solutions a.) To find v we can use the following formula: v v since the velocity is always to the velocity i.e. no tangential acceleration b.) To find R we can use the following formula: sin 1 R d using trig sin 1 d R c.) To find m q / we can use the following formula: qB mv R 1 1 sin sin Bd v d B v BR v m q 1 sin Bd v m q negative from the left hand rule (for negative particles) Right hand rule for positive particles d.) To find we can use the following formula: 2 d tan sin cos 1 sin cos 1 cos 1 1 1 2 d d d R R R d with more trig tan sin 1 1 2 d d d...
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hw6.11 - point where the particle exits the magnetic field...

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