Physics magnetic field question?
Please help. Im very confused Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 160 V/m and the magnetic field is 3.14×10−2 T . The ions next enter a uniform magnetic field of magnitude 1.72×10−2 T that is oriented perpendicular to their velocity. How fast are the ions moving when they emerge from the velocity selector? If the radius of the path of the ions in the second magnetic field is 17.2cm , what is their mass?
Strength of the magnetic field around a wire Physics Question?
A typical power transmission line carries a current of 1000 A about 30.0 meters above the ground. The magnetic field generated at ground level by the current is a matter of concern to those who live nearby. (a) What is the strength of the field at ground level below one such wire? (b) The strength of the Earth's magnetic field is about 5×10−5 T at sea level in middle latitudes. How close would you have to be to the power line to experience from it a magnetic field of equal strength? ** I solved (a) = .00000667. Not sure how to solve for (b). Please help. Thanks!**
Physics 2: Magnetic Field Between Two Wires?
1) Determine the magnitude of the magnetic field midway between two long straight wires 1.50cm apart when one wire carries 6.00A and the other carries 12.8A. Assume these currents are in the same direction. 2) Determine the magnitude of the magnetic field when these currents are in opposite directions. I cannot seem to get part 1 right, although I was able to get the second part right. For part 1 I figured B1-B2=Btotal B=( Uo*I)/(2*Pi*r) Calculated both B1 and B2, subtracted and got -1.813E-4 T and that is not correct. For part two I did the opposite, B1+B2=Bt= 5.01E-4T and that was correct. I'm not sure where my math is going wrong or If I am going about part 1 wrong, but I would appreciate any help with the calculations, I am down to my last two tries for the problem as well:/ Thanks
Physics 2 question about magnetic field. Help would be appreciated!?
you said it is in the figure but i didn't see any figure, so im just making up my own figure. You know if you move a charge in an electric field( in your case the two metal plates) +ve charge will experience a force in the direction of the field and -ve charge opposite to the field. So clearly, the direction of the magnetic force should be opposite (but equal in magnitude) to the electric field to pass in between the plates undeflected. So,to find the direction of the magnetic field, apply Fleming's Left Hand Rule(i suppose you could do that) Mathematically, Bqv=qE (letters have their usual meanings) Bv=V/d ( E=V/d) B=V/vd (here V=potential difference v =velocity of the charge d=separation between the metal plates) put the value and you get the answer
Can you answer this Physics JEE question from Magnetism?
The answer to the above question is a) ((μ i/(3r)) x 7/8) (k direction).The structure can be thought of as made up of four sections, an arc of reflex angle 300°, an arc of angle 60° and two straight lines pointing towards the center.These four sections contribute to the magnetic field at the center of the loop.An arc of angle θ produces a magnetic field of magnitude (μi/2r)x(θ/360) and direction which can be determined by where the thumb will be pointing if we curl our fingers in the direction of the current.A straight line produces 0 magnetic field at any point along its length.Thus, the magnetic field at the center of the loop will be(μi / 2 x 2r) x (300/360) pointing out of the plane by the arc of reflex angle 300°.(μi /2r) x (60/360 ) pointing out of the plant by the arc of angle 60°.Zero by the two straight segments.Therefore the total magnetic field = (μi / 4r)x(5/6) + (μi / 2r)x(1/6) + 0 + 0 = (μi /(3r)) x (7/8) out of the plane. As that is the direction of the z-axis, the direction of (k).So, The answer to the question is a) ((μ i/(3r)) x 7/8) (k direction).
Stuck on a Physics question...magnitude of magnetic field?
An electron which moves through a velocity selector (E=4.0kV/m and B=2.0mT) subsequently follows a circular path (radius=4.0mm) in a uniform magnetic field. What is the magnitude of this magnetic field?
Particles through a uniform magnetic field question for Physics II?
F = q(v x B) = qvBsinθ F = mv²/R = qvBsinθ q = mv/BRsinθ We are given that mv is the same for all particles, and B is the same as well. That means the value of R and θ determines then the value of q and its sign Particle 1: positive q, middle value q Particle 2: negative q, smallest value q Particle 3: q = 0 since no curvature Particle 4: negative q and largest value q since R is smallest
Physics, electricity and magnetic field?
If both currents are in the positive axis directions, then their magnetic fields will be in opposite directions and will therefore oppose each other. Assuming conventional current, the field will be OUT of the page from the x-axis current, and IN to the page from the y-axis current. Use the right-hand grip rule. (If electron current, use the left-hand rule which reverses the directions). Units not given for displacement, so metres assumed. Magnetic field at P(4,3) due to x-axis current (3m away from x-axis) is B = u0*7A/2pi*3 = 28pi*10^-7/6pi T = 4.67*10^-7 T OUT of page Magnetic field at P(4,3) due to y-axis current (4m away from y-axis) is B = u0*6A/2pi*4 = 24pi*10^-7/8pi T = 3.00*10^-7 T IN to page So net B field at P(4,3) is: (4.67 - 3.00)*10^-7 T = 1.67*10^-7 T out of page, perpendicular to y and x axes