Question

Determine the magnitude of the force on an electron traveling 6.55×10^5 m/s horizontally to the east in a vertically upward magnetic field of strength 0.40 T. Determine the direction of the force on an electron. 

          Determine the magnitude of the force on an electron traveling 6.55×10^5 m/s horizontally to the east in a vertically upward magnetic field of strength 0.40 T. Determine the direction of the force on an electron.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Determine the magnitude of the force on an electron traveling 6.55×10^5 m/s horizontally to the east in a vertically upward magnetic field of strength 0.40 T. Determine the direction of the force on an electron.
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Transcript

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00:01 Here for the solution, this is the free body diagrams, this is the b electrons, the v and there are the xxx, y -axis and x -axis, and to solve the next step.
00:18 Here, we know the equation that the force on electron is equal to f vector equal to qv multiplied by b.
00:26 Here we know the equation that force on electron f vector equal to q v multiply by b and here we substitute all the values in the equation here the value of q is 1 .6 multiply by 10 to the power minus 19 the value of v is 8 .75 multiply by 10 to the power minus 19 the value of v is 8 .75 multiplied by 10 to the power 5 i cap and for the b is 0 .45 j cap and by solving this we get f equal to 6 .3 sorry 6 .3 multiply by 10 to the power minus 14 k 6 .3 multiply by 10 to the power minus 14 k k cap newton so this is the force of on electrons.
01:53 Here, the direction of the force is towards minus along with the xxas.
02:01 Here in this, the direction of force is towards minus along the xx.
02:27 So this is the solution...
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