Question

The figure below shows a top view of a bar that can slide on two frictionless rails. The resistor is

R = 5.00 Ω,

and a 2.50-T magnetic field is directed perpendicularly downward, into the page. Let ℓ = 1.20 m.

A vertical bar and two parallel horizontal rails lie in the plane of the page, in a region of uniform magnetic field, vector B_in, pointing into the page. The parallel rails run from left to right, with one a distance ℓ above the other. The left ends of the rails are connected by a vertical wire containing a resistor R. The vertical bar lies across the rails to the right of the wire. Force vector F_app points from the bar toward the right.

(a) Calculate the applied force required to move the bar to the right at a constant speed of 2.50 m/s.
  N (to the right)

(b) At what rate is energy delivered to the resistor?
W

Answer 1

1. The above problem is related to force on a current carrying wire because of magnetic field.
2. All the steps are clearly mentioned and easy to understand.
3. Hope the answer will help you. Thank you for your support.
4. 
   x 100 im The magnetic X "app x x solution ! (a) Given the resistance of the registar R-5.00 u field •R l that directed pu perpendicularly downwand into the paper B = 2.50 T The length of the bar l = 1.20 m and the speed of the bar v= 2.50 m/s Now, The applied force required to move the bar FB Il éx8') = IB Simo where the current of the wire in the circuit l is the length of Bar. the angle between wine and magnetic field. Here, magnatic field is perpendicular downwards into that is o=90 F = Ils 1 is is the paper
   
   Now I - LEI is the equation for emf induced in the bar E = - Blue where B is the magnetic field I is the length of the bari is the velocity of the bar So the Induced Current |- Blol R R The ne the current Counterclockwise (in out of the page current flow upward through the bar, the magnetic force on the bar is to the left, so applied force must be to the the right The work done by this applied force Fida the power input from. the force sign indicates L So Our . So 1 Fidx dt All of by this power must be dissipated the resistor so the current is P = IR Fu I = R VR 34 From a 2 En 2 and en 4 we get 1- Blael = R VR fil Fu R RE
   
   5.00 F = B²l² u (2.50) ²x (1.20) ²x2.50 R F = 4.5 N applied force (6) The rate at which energy delivered to the resistor IP = Fu = 4.5x2.50 = 41.25 W