Recent content by EvanQ

alright got it now, thanks a lot for your help. i was about 80% through typing out my reworking to see if you could check it when i accidentally clicked the back button on my mouse and lost it all, and i really can't put myself through typing it again lol.

Homework Statement Three point charges lie along the x axis. A positive charge q1=16.0mC is at x=2.00m. Another positive charge q2=9.00mC is at the origin. Where should we put a third charge so the resultant force acting on it is zero? What should be the sign of the charge? Homework...

really confused sorry :(

9.8Msin180=ILBsin90??

and the current would go to the left, determined by the right hand rule?

ok so force pulling the wire down: F=masinθ = -9.8Msinθ so for the rest: F=ILBsinθ 9.8Msinθ=ILBsinθ 9.8M=ILB I=9.8M/LB ??

Homework Statement A straight piece of conducting wire with mass M and length L is placed on a frictionless incline tilted at an angle theta from the horizontal. There is a uniform, vertical magnetic field vecB at all points (produced by an arrangement of magnets not shown in the figure)...

ΔB = μo/4π(IΔLsinθ)/R^2 so I = 27 θ = 45 ΔL = 2mm? and R can be found using trig. so what is μ and o?? soz just really lost. any help would be great.

Homework Statement A wire carrying a 27.0A current bends through a right angle. Consider two 2.00mm segments of wire, each 3.00cm from the bend. Find the magnitude of the magnetic field these two segments produce at point P, which is midway between them. Homework Equations F =...

that was correct :) thanks heaps for your help

r = mv/qb 0.008276 = (1.67x10^-27 x 1200) / (1.602×10^−19 x b) (1.602×10^−19 x b) = (1.67x10^-27 x 1200) / 0.008276) (1.602×10^−19 x b) = 2.42146x10^-22 b = 1.5115x10^-3

would i be correct in saying that using the arc length formula: pi/2 radians is equivalent to 1.3cm θ = s / r r = s / θ = 1.3 / pi/2 r = 0.8276 cm t = 0.013m / 1200m/s = 1.083x10^-5 s then w = θ / t = (pi/2) / 1.083x10^-5 = 145041.2121 w = w0 + at 145041.2121 = 0...

yeh i was about to ask how you knew the radius. ok i'll work on those and get back to you in a few mintues. **digs deep into the memory bank for equations**

i got 1.129x10^-13 and failed my first attempt.

ok so: F = qv x B B = F/qv = F / (1.602 × 10^−19) x 1.2 (charge of a proton x velocity of proton) F = ma? = 1.67x10^-27 x a (mass of a proton x acceleration) however, as: m(v/t) / q(v) = m(1/t) / q = (1.67x10^-27 x (1/t)) / 1.602 × 10^−19 (where t = distance / velocity??)