Please answer all questions. Thank you! 5, + -15 points OSColPhys2016 22.11.PO73

Question

Please answer all questions. Thank you!

5, + -15 points OSColPhys2016 22.11.PO73. My Notes Ask Your Consider the figure. (a) A pendulum is set up so that its bob (a thin copper disk) swings between the poles of a permanent magnet as shown in the figure. What is the magnitude (in N) and direction of the magnetic force on the bob at the lowest point in its path, if it has a positive 0.210 c charge and is released from a height of 21.0 cm above its lowest point? The magnetic field strength is 1.40 T. (Assume the +x-axis is in the direction of B and the ty-axis is up along the page.) magnitude direction (b) What is the acceleration (in m/s2) due to the magnetic force of the bob at the bottom of its swing if its mass is 24.0 grams and it is hung from a flexible string? m/s2 magnitude direction

Explanation / Answer

Velocity at the lowest point can be calculated as

v = sqrt(2gh)

v = sqrt(2*9.8*0.21)

v = 2.0287 m/s

Now, The pendulum will experience magnetic force given by

F = qvB

F = 0.210e-6*2.0287*1.40

F = 5.964e-7 Newtons (in the upward direction)

Now, to find acceleration, use newton's second law

F = ma

a = F/m

a = 5.964e-7 / 0.024

a = 2.485e-5 m/s2

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