Asked by rick
There is a 0.225 T magnetic field directed along the +x axis and a field of unknown magnitude along the +y axis. A particle carrying a charge of 7.20 10-5 C experiences a maximum force of 0.455 N when traveling at a speed of 2 104 m/s through the region where the fields are. Find the magnitude of the unknown field.
Answers
Answered by
Damon
Do you mean V = 2*10^4 ???
F = q ( V x B )
the force is maximum when the velocity vector is perpendicular to the B fieldso consider the B field in the x y plane and V in the z direction.
then
|F| = |q| |V| |B|
so
.455 = 7.210^-5 * 2 * 10^4 * |B|
|B| = .316 T
so
Bx^2 + By^2 = |B|^2 = 0.1 T
.225^2 + By^2 = .1
.05 + By^2 = .1
so By is also about .22 T
F = q ( V x B )
the force is maximum when the velocity vector is perpendicular to the B fieldso consider the B field in the x y plane and V in the z direction.
then
|F| = |q| |V| |B|
so
.455 = 7.210^-5 * 2 * 10^4 * |B|
|B| = .316 T
so
Bx^2 + By^2 = |B|^2 = 0.1 T
.225^2 + By^2 = .1
.05 + By^2 = .1
so By is also about .22 T
Answered by
sandra
where did you get .1 from?
Answered by
Quinn
The .1 is from |B|^2 which is equal to .316^2
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