If you are looking downward from the positive y direction, what is the expected

Question

If you are looking downward from the positive y direction, what is the expected direction of rotation of the coil? A particle with positive charge q = 6.41 times 10^-19 C moves with a velocity v = (3i + 4j - k) m/s through a region where both a uniform magnetic field and a uniform electric field exist. What is the equation for the force exerted by a magnetic field on a moving charge? N What angle does the force vector make with the positive x axis? You appear to have correctly calculated this angle using your incorrect results from part (a).^o

Explanation / Answer

v = 3i + 4j - k

B = 5i + 5j + k

E = 2i - j - 5k

electric force, Fe = qE = 6.41e-19 ( 2i - j - 5k)

magnetic force, Fm = q( v X B)

= 6.41 e-19 [ (3i + 4j - k ) X ( 5i + 5j + k ) ]

= 6.41 e-19 ( 15k -3j - 20k + 4i - 5j + 5i )

= 6.41e-19(9i - 8j - 5k) N

F = Fm + Fe

Fx = 6.41e-19 ( 9 + 2) = 7.05e-19 N

Fy = 6.41e-19 ( -1-8) = -6.07e-19 N

Fz = 6.41e-19(-5-5) = -6.41e-19 N

b) to find angle we will use dot product.

F.i = |F||i| cos@

(7.05i - 6.07j -6.41k)e-19 . (i) = (sqrt(7.05^2 + 6.07^2 + 6.41^2))(1) cos@

7.05 = 11.30 cos@

cos@ = 0.624

@ = 51.40 deg

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