A proton moves with a velocity of v = (2i - 4j + k)m/s in a region in which the magnetic field is B = (i + 2j - 3k)T. What is the magnitude of the magnetic force this particle experiences?
2007-11-30 04:46:57 · 3 answers · asked by andrew w 1 in Science & Mathematics ➔ Physics
Another way to evaluate v X B is to remember the unit vector circle for cross products. Following around the X clockwise is positive; going around it CCW is negative.
i X j
. k
Example: j x k = i
but: i x k = ( - j)
(2i - 4j + k) x (i + 2j - 3k)= [(2ix2j)+ (2i)x(- 3k)] +[(- 4j)x i +(- 4j)x- (3k)] +[k x i +k x 2j]
v xB = 4k + 6j +4k -12i +j -2i
= -14i +6j +8k
mag(vXB) = sqrt (14^2 + 6^2 + 8^2)
=sqrt(296) = 17.2
F= q (v X B) = 27.527 x 10^-19 Newtons
where q = +1.6x10^-19 Coul
2007-11-30 05:29:14 · answer #1 · answered by Anonymous · 2⤊ 0⤋
The force is q(vxB) where q is the charge on the proton, v is the velocity, and B is the magnetic field. All you have to do is find the cross product, vxB and mutiply it by the charge of the proton to get the force vector. The magnitude will be the length of that vector.
The previous post made a mistake in the computation of the cross product. The coefficient of k should be 8, not 16.
2007-11-30 04:59:35 · answer #2 · answered by mathematician 7 · 0⤊ 2⤋
F = q* v X B where q = 1.6x10^-19 Coul, v and B are given and the "X" means cros product
v X B = (2i -4j +k)X(i + 2j - 3k) = i*(12 -2) +j*(1 -(-6))+k*(4-(-4))
= 10i + 7j +16 k T-m/s
you can work out F
2007-11-30 04:57:37 · answer #3 · answered by nyphdinmd 7 · 0⤊ 2⤋