Question

1.) We consider a thin metal strip with conductivity o and thickness d oriented in the (x,y)-plane and moving at constant velocity v = vj in the y-direction, as shown. The metal strip falls in the air gap between the rectangular pieces of an electromagnet. The cross-sectional with of each pole is w along the r-direction) and length 1 (along the vertical). The electromagnet produces the magnetic field B = Bk (out-of-the-page). We treat this problem approximately by considering the analog of a battery of electromotive force and internal resistance r that is connected in series to an external resistive load R. We assume that the metal strip is much larger than the cross-sectional area of the electro- magnet; this means that we can ignore edge effects of the metal strip. a) Describe an equivalent circuit with battery E, internal resistance r, and resistance R of the entire strip. We ignore the calculation of R. b) Calculate the drag force!

Answer 1

Answer:

a)

Rh MW k, Shunt Here we are considering the analoge battery of the traf Celectromotive force) a & Internal resistance & and connected to a load Eternal resistance Raby ignore the calculation of R , let us determine the internal resistance of a cell. Potentiometer is a device used to measure the internal resistance of a call to compare the ent two cells and potentiometed potential difference across a resistor. (page-1)

It consists of a long wire of uniform cross sectional area and of min length. The matenol of wire should have a high resistivity and low temperature centrent. The wires are stretched parallel to each other on a wooden board. The wires are joined in senes by using thick metal strups - A meter scale is also tached on the wooden board. It works on the principle that when a constant cument flows through a wire of uniform Gross sectional area , potential difference between. It's two points is directly proportional to the length of the wire between the two points. → Relation between Ent (e), intend restorce (8) and potential difference of cello . If a cell emf CE) and internal resistance (1), connected in senes with an external R etaoce (R), then the curait has the total resistance (R+r). The current in the circuit is given by I = { tance Rto page-

Rate of flow of electrons will depend upon drift velocity and drift velocity is inversely proportional to area of cross-section, v_d ∝ 1/A. Therefore R ∝ 1 / A.

of € - I ( RD) Hence, v = IR = &- Ir This means, v is Dess than Eby an arrow equal to the fall of potential inside the cell due to its internal resistance → Forom the above equation, re-V V The internal resistance of the cell r=RE-V > using a potentiometer, we can adjust the Theostat to obtain the balancing lengths hy and Iz to the potentiometer for open and clos cucuits respectively Then & = kl, and v= klei where, K is the potential gradient dong the wiso. Now we can modify the equation for getting the Penternal resistance of the given cell, by using the above relations are [page-3

(b)

The drag force, F_D, depends on the density of the fluid, the upstream velocity, and the size, shape, and orientation of the body, among other things. One way to express this is by means of the drag equation. The drag equation is a formula used to calculate the drag force experienced by an object due to movement.

F_D = C_DA ρV²/2,

Where,

F_D is the drag force,

C_D is the drag coefficient

A is the reference area

ρ is the density of the fluid,

V is the velocity relative to the object.

The reference area, A, is defined as the area of the orthographic projection of the object on a plane perpendicular to the direction of motion. For hollow objects, the reference area may be significantly larger than the cross-sectional area, but for non-hollow objects, it is exactly the same as a cross-sectional area.