A long, straight, vertical wire carries a current upward. Due east of this wire, in what direction does the magnetic field point?
... using the right hand rule for current in a conductor I get north When you get a good response, please consider giving a best answer. This is the only reward we get.
Two long, straight wires carry currents in the directions shown?
http://gauss.vaniercollege.qc.ca/webwork2_course_files/NYBqi/tmp/gif/bou345-637-setNYBset01-ch30-prob2--../Graphics/ch-30-e03.gif Two long, straight wires carry currents in the directions shown in the figure above. I1 = 4 A, I2 = 12 A, a = 6 cm, b = 2 cm. a) Find the magnetic field strength at point P. B = ( ? i + ? j ) T b) What force would 1 m of a third wire, carry a 2-A current out of the page, experience if place at P? F = ( ?i + ?j )N
N the figure below, the current in the long, straight wire is I1 = 8.00 A and the wire lies in the plane of th?
The forces on the 2 horizontal parts of the loop in your diagram would yield zero. Find the magnetic field at the vertical parts of the loop due to I1. By Ampere's law, ∫ B ▪ dL = (µo)(I1) (Bl)(2πc) = (µo)(I1) and (Br)(2π(c + a)) = (µo)(I1) (left and right portions, respectively) (Bl) = (µo)(I1)/(2πc) and (Br) = (µo)(I1)/(2π(c + a)) Find the forces on the vertical parts of the loop. Since the currents are in opposite directions, the forces will also be opposite in direction. F = (I2)(L)(Bl) - (I2)(L)(Br) F = (I2)(L)(µo)(I1)/(2πc) - (I2)(L)(µo)(I1)/(2π(c + a)) F = [(I2)(L)(µo)(I1) / (2π)] [1/c - 1/(c + a)] F = [(10.0 A)(0.250 m)(1.257e-6 N/A²)(8.00 A) / (2π)] [1/(0.100 m) - 1/(0.100 + 0.150 m)] F = 2.40e-5 N
Two circular loops of wire with a common center?
i apologize if this isn't right, but in the audio world, you can minimize the effect of placing coils near each other by placing them perpendicular to each other, so i think the field in the center will be lower than either field separately because the lines of force cancel out. my guess would be either zero, or some very tiny number. sorry i can't give a straight number answer, but i hope it a least gave you some direction for the calculations maybe.
A. Find the magnitude and direction of the force that the magnetic field of wire 1 applies to a 1.5-m section?
Currents flowing in opposite directions cause repelling forces on the two wires (the circular field lines from each current run in opposite directions between the two wires .. causing the field lines to be 'compressed' in this region .. resulting in the sources being repelled) Field created by i1 at distance d = 0.065m .. B1 = μₒ.(i1)/2πd B1 = (4π^-7)(15A)/ (2π.0.065m) .. .. B1 = 4.62^-5 T Force on i2 passing perp. through B1 .. F = B1.i2.L F = (4.62^-5 T)(7A)(1.50m) .. .. .. ►F = 4.85^-4 N (away from i1)
What are the magnitude and the direction?
Break it down to what is given first: μ0 = 4(pi) x 10^-7 (permeability constant in a vacuum) I = 2.50 A r = 4.50 cm = 0.045 m v = 6.00 x 10^4 (I think that's what you wrote as the velocity but i'm not sure because your formatting was confusing. If this isn't what the velocity was, then use my same method but just plug in your own value for v) Θ = 90 degrees (because the electron is moving directly towards the wire, which would imply a perpendicular motion to the current) q= charge of electron = 1.6x10^-19 B= (μ0)(I) / (2)(pi)(r) Plug in your values: B = (4pi x 10^-7)(2.50) / (2)(pi)(0.045) = 1.11111111... x 10^-5 (I stored this answer in my calculator as A) F= Bqv(sinΘ) F = (A)(1.6x10^-19)(6x10^4)(1) F = 1.06666667 x 10^-19 N ^ That is the magnitude of the force The direction of the magnetic force relative to the direction of current is perpendicular; meaning, a direction of 90 degrees.
Current in a loop of a half circle?
The field is one-half the field at the center of a full circular loop of 4.25 m radius. B = mu0*I/(4R) = 1.4783965E-7 T The center of the circular sections is at the center of the vertical straight wire, where the circular field due to that wire = 0, since enclosed current along the wire's centerline = 0. For similar reasons the two horizontal straight sections contribute nothing to the field at the center point. They have zero axially-directed fields, and zero circular fields along their centerlines (and the extended centerlines). With the quarter-circle sections, each length increment ΔL on one section has a corresponding ΔL exactly opposite it on the other section such that their fields cancel at the center point. Cancellation happens because the currents in each ΔL have the same "inertial" direction, and clockwise fields around each when viewed along this direction, so at the center one field is directed out of the plane and the other is of equal magnitude and directed into the plane. This leaves only the large semicircle contributing to the field. See the refs. which explain the derivation of the one-loop field and the value of the half-loop field.
The force on a wire is a maximum of 5.30 N when placed...?
The force on a wire is a maximum of 5.30 N when placed between the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on. a. what type of magnetic pole is the top pole face? b. if the pole faces have a diameter of 10.0 cm, estimate the current in the wire if the field is 0.15 T. c. if the wire is tipped so that it now makes an angle of 10 degrees with the horizontal, what force will it now feel? Thanks for any help!
Please help me with this magnetism question?
I'm going to assume that the currents are flowing in the positive x and positive y directions... a) Use the right hand rule. The x wire's field at point P goes out of the page, and the y-wire field goes into the page. Field due to x wire: B = ui/2πr = 2E-7i/r = 2E-7*5/0.4 = 2.5E-6 Field due to y wire: B = 2E-7*3/0.3 = 2E-6 x wins. Net magnitude: 2.5E-6 - 2E-6 = 5E-7 T Since x wins, the field goes out of the page. b) I'm going to call that point Z Distance from x wire to point Z: √(0.31²+0.4²) = 0.506 m Plug into formula B = 2E-7*5/0.506 = 1.98E-6 T The direction is a little bit tricky. It will be pointing out of the page and towards you. I'm going to call the positive x axis east, positive y axis north, and out of the page up. So the angle would be arctan(0.4/0.31) = 52. [S 52 UP]. Just draw a picture.. it's hard to describe how I got that Do the same thing with y. √(0.31²+0.3²) = 0.431 m B = 2E-7*3/0.431 = 1.39E-6 T angle: arctan(0.3/0.31) = 44. [E 44 DOWN] So now combine 1.98E-6 [S52U] and 1.39E-6 [E44D] using components. In the vertical direction: 5.94E-7 up In the horizontal direction: 1.58E-6 [E51S] net electric field: 1.69E-6 T [E51S21U] uhh. yeah. I'm not sure if that's right. Someone else might have an easier way of doing this..