The drawing shows a wire composed of three segments, AB, BC, and CD. There is a current ofI= 3.1Ain the wire. There is also a magnetic fieldB(magnitude =0.74T) that is the same everywhere and points in the direction of+zaxis. The lengths of the wire segments areL_AB= 1.63m,L_BC= 0.48m, andL_CD= 0.7m. Find the magnitude of the TOTAL force on the wire.

+z B C +y +XGiven that, the lengths of the wire segments are LAB = 1.63 m + 2 LBC = 0148m D B Lep = 0 0 7 m B +Y magnitude of magnetle A + x field, B= 0.74T intz direction. Current in the wore, 1= 301 A we know ... if the current cuts across the magnetic freld al- on anagle. A the force on current is expressed ass ! 'F = BX IXLX SMO ( 0 is the angle between length element and B).. The force exerted on the AB Sayment of the wire is, FAB =. BILAB SME The diagram shows the angle between AB length element and magnetic field ( B ) is 02 90 0 - : FAB = ( 0 . 7 4 X 3 . 1 x 1 . 6 3 x Sim 90 0 ) N 3.74 N According to Fleming's left- hand rule, the drove- tion of the force will be on the positive 4-axis The force exerted on the BC Segment of the wire is Fine = BILpre Simo Here, siso the angle Between Be length and B is 0 2 90' so , Fore = (0 . 74 x 3. 1 x 0 . 48 x sin 90' ) N 2 10l N According to the Fleming's left-hand rule, the direction of the force will be on the negative x- axis. The diagram shows the angle between the colengit , element and the magnetic field, 020. 40, FCD = BILCD SMO = ON ": The magnitude of the total force on the wise 18 , F = ( FAB ) ?+ ( Fme ) 2+ ( FCD ) 2 . 2 / ( 3 . 74) 27 ( 101 ) 2 2 13.9 N Ans