Magnetic Force on a Triangular Loop?
The triangular loop of wire shown in the drawing carries a current of 4.79 A. Angle W is 43 degrees. The length of side AB is 1.9 m. A uniform magnetic field is directed parallel to side AB and has a magnitude of 1.97 T. Calculate the magnitude of the magnetic force acting on each side of the triangle. Diagram*: http://www.cramster.com/answers-mar-09/physics/magnetic-force-the-triangular-loop-of-wire-shown-in-the-drawing-carries-a_510420.aspx *The link to the diagram has a problem with different numbers for the current, angle, and magnetic field. Please refer to my numbers, thanks! The answers that I got were: F(AB) = 0.00 N F(BC) = 16.719 N F(AC) = 16.719 N Can anyone verify that my numbers are correct? Also, should any of the forces be negative? I know that the net force on the triangular loop is 0.00 N. Thanks a lot.
The triangular loop of wire shown in the figure carries a current 5.00 in the direction shown. The loop is in?
F = BIL sin θ ======================== The magnitude of the force exerted by the magnetic field on side PQ of the triangle, = BIL sin 0 = 0 =============================== RQ = √ (0.6^2 + 0.8^2) = 1 Angle between B and I is found from sin θ = 0.8/1.= 0.8 The magnitude of the force exerted by the magnetic field on side RQ of the triangle, = BIL sin θ =3*5*1*0.8 = 12 N ===================================== the magnitude of the force exerted by the magnetic field on side PR of the triangle, = BIL sin θ =3*5*0.8 sin 90 = 12 N ==================================== Resultant force ( both in the same direction ) is 12+ 12 = 24 N =================================
A right-triangular current loop carries a current I, and a constant magnetic field B is directed perpendicular to one edge of the loop.?
A right-triangular current loop carries a current I, and a constant magnetic field B is directed perpendicular to one edge of the loop (see figure below). If θ = 38°, L = 50 cm, and B = 2.6 T, what is the total force on the loop? Magnitude:_____N Direction:
The triangular loop of wire shown in the drawing carries a current?
For each side, F = IL X B, a vector cross product, and |F| = ILBsinθ, where θ is the angle between IL and B. To solve the problem, for each side use trig to get the length L and simple geometry to determine θ. Then plug the values into the equation. Note θ = 0 for side AB so its force = 0. And the signs of the forces on AC and CB are opposite; one is into the page and one is out. I'll leave it to you to determine which is which; use the right-hand rule (ref.). And the magnitudes of the forces on AC and CB are equal (|BCsinθ| = |AC|), so ... answer B = 0!
What is the magnitude of magnetic field at the centre of a hexagonal loop of a metallic wire having each side 'a' and carrying current I (μI√3/Πa)?
Suppose that the current I flows through the hexagonal loop ABCDEF of each side= a.The magnetic field produced at the point O due to each of the six sides of the hexagonal loop is same and perpendicular to its plane (in outward direction).Here is the solution:—If perpendicular distance of the point O from the side AB (that is OM=r), then magnetic field at the point O due to the side AB of the loop= B(AB)The magnetic field due to the whole hexagonal loop, B (loop)= 6 x B(AB)B (loop) = √3μI/πaTushar Raj
Physics: Magnetic Fields?
A loop of wire carrying a current of 2.0 A is in the shape of a right triangle with two equal sides, each 15 cm long. A 0.7 T uniform magnetic field is in the plane of the triangle and is perpendicular to the hypotenuse. The resultant magnetic force on the two sides has a magnitude of: A) 0 B) 0.21 N C) 0.30 N D) 0.41 N E) 0.51 N The answer key says C. Can someone please explain why. I got B. Thanks.
How do we find the magnetic flux through a square loop placed at a distance 'a' of an infinite wire which carries a current I which changes with time?
Do not worry about the time dependence part.Write forget i (t)Treat it as I.Now…take infinitesimal portion of area…at a dist x from closer sideArea =a.dx (a…side length)B=K i/2pi (x+d)*d= dist from wireNow small flux = B.dxCalculating integral ..u get a formFlux=[ln (a+b/b)][(K)Ia/2pi]Now replace I by I (t)Now u can understand that you get a flux..chnging with timeFurther d (flux)/dt gives the emf induced
What is the magnetic flux near an infinitely long current carrying wire in a triangular region near it?
What do you mean by “ in a triangular region near it?”Edit DraftRequestFollow3Comment1Share
Magnetic Field Question!?
a) When the triangle crosses y+, magnetic fields start to flow through the triangular loop. The loop starts to generate current such that it would resist the increase of magnetic field flow. Therefore, an induced current would flow counterclockwise. b) When the loop crosses the -x axis, there will be little change in magnetic field flow through the loop. So no current would be induced. c) When the loop crosses the -y axis, the magnetic field flow through the loop will start to decrease. Therefore, a current would be induced in the loop to resist the change. An induced current in the clockwise direction will compensate the decrease. d) There is no magnetic field in the region right to the y axis, so nothing happens to the loop. No induced current when the loop crosses the +x axis.
Magnetic force exerted on the triangle...?
F (mag) = I [ L cross B] = BIL sin x p = angle between B direction and length (vector) ------------------------------------ side AB >> F(AB) =0, sin p =sin 0 =0 -------------------------------------- along AC >> L = 2 tan 55 = 2.856 m p = 90 F(Ac) = 1.8*4.70*2.856* sin90 = 24.16 N direction> out of screen (normal) ------------------------- along CB >> L = [2^2 + 2.856^2]^1/2 = 3.487 m p = 55 (mark alternate angle in AB || B direction) F(CB) = 1.8*4.70*3.487* sin55 = 24.16 N direction> into the screen (normal) No net force on triangle >>> equal & opposite forces constitute couple.