Fuzhou photography v2

(1) When a square wire frame passes through a uniform magnetic field, let the induced electromotive force on the edge of cd be E, the current intensity in the wire frame be I, and the voltage between C and D be Ucd.

E=BLv

From ohm's law, I = er?

Ucd=34IR

The solution is Ucd=0.45V v v.

(2) When the square wire frame passes through the magnetic field area at a uniform speed, let Ampere Force be F and Zhang Liwei T on the thin wire be, then

F=BIL

T=m2gsinθ

m 1g=T+F

During the movement of the square wire frame before it enters the magnetic field, according to the conservation of energy, then

m 1gh？ m2 GHS inθ= 12(m 1+m2)v2

The solution is m 1=0.032kg, m2=0.0 16kg?

(3) Because the acceleration of the wire frame moving in the magnetic field is the same as before (only under the action of gravity), the total mechanical energy of the wire frame and the object P remains unchanged in the process of passing through the magnetic field area, so the work W done by the force F is equal to the Joule heat Q generated by the whole wire frame, that is,

W=Q

Let the Joule heat generated by the cd edge of the wireframe be Qcd, which is obtained according to Q=I2Rt.

Qcd= 14Q

The solution is Qcd=0.0575J j.

Answer: (1) When the cd edge of the wireframe moves in a uniform magnetic field, the voltage between C and D is 0.45V V V.

(2) The mass m 1 of the wireframe and the mass m2 of the object P are respectively m 1=0.032kg, and m2=0.0 16kg?

(3) The Joule heat generated by the cd edge of the square wireframe is 0.0575J J.