DO PHYSCS ONLNE MOTORS AND GENERATORS MAGNETC ORCES ON CURRENTS There are magnetic forces on moving charge particles. ut, a moving charge correspons to an electric current. Therefore, a conuctor carrying a current in a magnetic fiel will experience a force. t is the force on a conuctor carrying a current that is responsible for the rotational motion of an electric motor, the vibration of a membrane in a louspeaker an the eflection of a neele in an analogue electrical meter. This is often calle the motor effect. Consier a straight conuctor of length L carrying a current in a uniform - fiel of strength. Then, the conuctor will experience a force. The irection of the force is etermine by the right han palm rule an its magnitue is given by equation () () Lsin where is the angle between the conuctor (current) an the irection of the -fiel as shown in figure (). conuctor uniform magnetic fiel Lsin thumb L L palm: force into page fingers 0 0 uniform magnetic fiel irecte into page ig.. The irection of the force on a current is etermine by right han palm rule. The principle of a force on a current in a magnetic fiel is illustrate in how a louspeaker works. An auio signal is connecte to the wire leas of a speaker which are connecte to a coil that is attache to the speaker cone. The speaker
cone is mounte so that it is free to vibrate back an forth. A permanent magnet is aligne with the coil in such a way that when a current passes through the coil, the force on the conuctor can move the speaker cone in or out. When an alternating current of the auio signal is connecte to the coil, the force on the conuctor alternates at the frequency of the auio signal. The speaker cone vibrates proucing a series of compressions an rarefactions of the ajacent air, thus proucing the soun wave with frequencies an intensities to give a reprouction of the electrical signal an the original soun signal. The main features of a louspeaker are shown in figure (). coil wining attache to speaker cone N -fiel current through coils S force on cone N permanent magnet speaker cone rigi frame attache to magnet an speaker cone ig.. Louspeaker. When a AC current passes through the coil, an alternating force moves the speaker back an forth to prouce the soun wave. An any instant, the irection of the force on the speaker can be etermine from the right han palm rule. v q right han palm rule
Magnetic force between two parallel current carrying conuctors A conuctor carrying a current has a magnetic fiel surrouning it. A conuctor carrying a current in a magnetic fiel experience a force. Hence, two current parallel carrying conuctors will exert a force on each other. igure 3 shows the force acting on conucting (current ) ue to the magnetic fiel surrouning conuctor (current ). The irection of the force is given by the right han rule. When the currents are in the same irection, the conuctors are attracte to each other. currents an are into page force on conuctor ue to conuctor think of two railway tracks merging in the istance currents in the same irection attract ig.. Two parallel conuctors carrying currents in the same irection are attracte to each other. When the currents are in opposite irections, the conuctors repel each other as shown in figure (3). 3
current out of page an are into page force on conuctor ue to conuctor ig. 3. Two parallel conuctors carrying currents in opposite irections are repelle to each other. At the location of conuctor, the -fiel from conuctor is o The magnitue of the on force conuctor in this -fiel is L o L Hence, we can conclue, that the force per unit length between two parallel conuctors is o o () k k o 4 0 T.m.A L 7 - The right han palm rule can be use to etermine the irection of the force between the conuctors. Example There are two parallel horizontal conuctors. The first conuctor has a current of 00 A. The secon conuctor lies 350 mm irectly below the first conuctor. The secon conuctor has a mass of 0.50 g.m -. What is the minimum current through the secon conuctor so that it oes not fall ue to gravity? What are the irections of the currents? 4
Solution How to a approach the problem Draw a iagram of the situation: labelle with known an unknown quantities Type of problem: force between parallel conuctors: o Magnetic force: L Weight : mg G or the conuctor to be balance, the gravitational force must equal the magnetic force an the conuctors must attract each other. Therefore, the currents an must be in the same irection. = 00 A = 0.350 m currents in same irection attract =? A G = mg = (m/l) g L g = 9.8 m.s - m/l = 0.50 g.m - =.50 0-4 kg.m - o = 4 0-7 T.m.A - o / = 0-7 T.m.A - Gravitational force: / / G m g G L m L g o Magnetic force: / L orces balance to support n conuctor o m / L g We can rearrange to fin the current 4 m / L g (.5 0 ) 9.8 0.35 A 5.8 A /.0 0 00 7 o P60 P609 P676 P638 P646 P6464 P648 P6547 P6576 P66 P6788 P6859 5