University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2022
Views:

127

Pages:

10
Author:

FrextWhestembki

Magnetic Force on current-carrying wire The force on this current-carrying wire is essentially the force on single charge carrier times the number of those carriers: 1) Magnetic force on single charged particle: F = qv  B 2) Number of charge carries: volume × n = LA × n 3) Total force on all carriers: Recall (Ch 25): I remember : carrier density n = (number of charge carriers per volume) Ftot = LA n qv  B = LA j  B = IL  B Q  nqv A t Magnetic force on current-carrying wire: F = IL  B vector L is in direction of current         F               B I L           F = IL  B This expression applies only to a straight wire in a uniform magnetic field. What if the wire is not straight? B dF = I ds  B   I                          ds        dF         infinitesimal force dF on an infinitesimal ds of current-carrying wire Integrate over the part of the wire that is in the magnetic field region. F =  dF  F = I ds  B  Example: a wire carrying current I consists of a semicircle of radius R and two horizontal straight portions each of length L. It is in a region of constant magnetic field as shown. What is the net magnetic force on the wire? I         B         R         L L         y x         There is no magnetic force on the portions of the wire outside the magnetic field region. Straight sections: F = IL  B L  B, so F1 = F2 = ILB I y       F1       R       L         F2 B     L           x Semicircular section: infinitesimal force dF on an infinitesimal ds of current-carrying wire I y      F1 dF ds d      R      L         F2 B       L            x ds subtends angle from  to +d. Why did I call that angle  instead of ? infinitesimal force is Because we usually use  for the angle in the cross product. ds  B, so dF = I ds B. Arc length ds =R d. Finally, dF = I R d B. dF = I ds  B. y-component of F: dF = I R d B. dFy = I R d B sin I  Fy =  dFy y 0  Fy =  I R d B sin 0 dFy    F1 dF d ds   R    L         L            x  Fy = I R B  sind 0  Fy =  -I R B cos 0 Fy = 2 I R B    F2 B    Interesting—just the force on a straight horizontal wire of length 2R. Does symmetry give you Fx immediately? x-component of F: dFx = -I R d B cos  Fx = -  I R d B cos 0  Fx = -I R B  cosd 0  Fx = -  I R B sin 0 Fx = 0 I y    F1 dF ds d  dF x R    L         F2 B       L            x Total force: Fy,tot =F1 + F2 + Fy,arc Fy,tot = ILB + ILB + 2IRB Fy,tot = 2IB L + R  F = 2IB L + R  ˆj I y Fy         F1 dF ds F2 B         R         L L                 x We probably should write the force in vector form. Example: a semicircular closed loop of radius R carries current I. It is in a region of constant magnetic field as shown. What is the net magnetic force on the loop of wire? FC         B         R         y x    I             We have just calculated the force on the semicircular part in the previous example (current is flowing in the same direction). FC = 2 I R B