Recent content by Bradsteeves

That equals 2.85

Ok I tried that so its (I2)^2 on one side so it would be = to F(2pie x d)/Mo(2)(l) which still equalled 1.19X10^-7 so i don't know how or what way u used to get it

Ok so then F=Mo(2)(I2)^2(l)/2pie(d) plug in 2.85X10^-5 for F, plug in 4pieX10^-7 for Mo, plug in 4 for l and plug in 0.08 for d. I calculated and ended up with 1.19 X10^-7, the answer is supposed to be 1.18 and 2.36

Well all I can get from that is that it would be Mo2(I2)(I2)(l)/2pie(d) and the I2's would cancel out so ur left with F = Mo(2)(l)/2pie(d) but then how can i solve for I? I put in the values into that equation and did not end up with the correct answer

Um yes they told me but I am not sure how I am supposed to use that in the equation when i still don't know either of the actual currents. How am I supposed to set up the equation knowing one current is twice the other current?

Homework Statement The magnitude of the magnetic force between two conductors 4.00m long, separated by 8.00cm is, 2.85X10^-5 N. The current in one conductor is twice the current in the other. Calculate both currents. Given: FM = 2.85X10^-5 N l = 4.00m Mo = 4pie...

i got 0.0196 for the tension?

Homework Statement Two particles, separated by an enormous distance, approach each other. Each has an initial speed of 3.0 X 10^6 m/s. Calculate their minimum separation, assuming no deflection from their original path. The mass of a particle is 6.6 X 10^-27 kg. Particle charge = 3.2 X...

Homework Statement Two identical small spheres of mass 2.0g are fastened to the ends of an insulating thread of length 0.60m. The spheres are suspended by a hook in the ceiling from the centre of the thread. The spheres are given identical electric and hang in static equillibrium, with an...