Recent content by rosstheboss23

I actually found out what I did wrong. Here is the appropriate way to approach this problem B=N(4pi times 10^-7)(I) Plug in I and B and N= turns per meter...

Homework Statement The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of 4.0 T, and the current in the solenoid is 300 A. What is the number of turns per meter of length of the...

Thank you. I really appreciate your help.

Homework Statement I was wondering how would you determine the magnitude an external field needed to cancel two magnetic fields in two wires? In this instance the currents of the two wires are in opposite directions...so I know there is a repulsion force. The currents in both wires are equal...

I have added the question by editing my initial information. If anyone can help me it would be greatly appreciated.

Homework Statement Suppose that I1 = I2 = 20 A; currents are running opposite each other, and that the separation between the wires is 0.019 m. By applying an external magnetic field (created by a source other than the wires) it is possible to cancel the magnetic field at the position of each...

Alright here is what you need to do. Once this finished you can put it in the archives for other struggling students... In order to get the force you will need to use the electric field equation. E=kq/(distance). To get the distance you will need to do the following: convert the units to...

If you multiply intensity by the area of the sail the units cancel to leave watts which is Joules per a second. Then if you multiply the intensity by the area you will get 1400W considering it is 1m(squared). Since the aircraft is in space any force is going to propel it so all you have do from...

Ok. Thanks I just found out the idea with this one. KE=mgh so .4479J/[(.39kg)(9.8m/s(squared)]. So the answer would be .1172m. Thanks again for all your help.

Ok. Would the conservation of energy for angular momentum go like this mgh =.5(1/3)(m)(L(squared))(3.5)(squared) + mgh(final). I tried this and got (.39)(9.8)(.75)=.4479(Part a answer) +(.39)(9.8)(height above bottom pt.) and I got .6328m. I find the concept hard to understand. Is this thin rod...

Ok. I will do so but for the potential energy part of conservation I would still use mgh right?

Ok I got part A. Formula for I. For those interested is (1/3)(m)(L(squared)). This is for a rod rotating about one end. Part B is I am still stuck on. Any advice?

Ok thanks. I will look up the formula, but I am not sure if its in my textbook. Does everything else look reasonably good? Also I was wondering if you knew anything on part B. I really appreciate your help.

Try using the kinematics equations replacing x with (radians), v with (angular velocity) and a with angular acceleration. I will reply if with additional information if you follow up on this.

You can determine angles using SOH CAH TOA. From there you can use components to work out your answer more precisely.