Question

QUESTION 20 A wire carries a current of 1.25 A from point A to point B in a zigzag path shown by the solid line in a uniform magnetic field of 0.250 T as shown below. What is the magnitude of the total force on the wire? 35.0° 50 cm 20 cm OA 0.174 N ОВ. 0.303 N Ос None of these answers OD. 0.219 N ОЕ. 0.125 N

Answer 1

We know the force acting on a current canning conductor of length e in a magnetic Biele FTXB wherre 1 is the current carrjes the conductore. bay M For triangle AOMB MB Sin 35° OB 50 CM 35° MB - 0B Sin 35° 0.5 Sin 35 m 20 cm 350 "OB= 50 cm = 0.5 m N A om and cos 35° OB - ОМ = OB COS 35° 0.5 COD 350 m triangle OA = 20 cm -0.2 m Fort AONA AN Sin 35 OA AN = OA Sin 35° 0.2 sin 350 m COS 35° ON OA ON = 0 ACOS 35° 0.2 cos 35 m as The Wirce cancies a current 1 = 1:25 A from point A to point B in the zigzag path. So we can write the length th of the wire l AN (-1) + NM + MB (-2) = (AN+ MB) (-2) + 'Not om

The uniform magnetic field is giv mazuletic 1 = 0.2501-7NT given by Now , ex B LX EK J X 0.2501 ol-7 g 950 + Tom [(AN+ MB) (-2) + (no tom) | | (An + MB) x (0-250) | 0:2 Sin95° +0.58in 35) X0.250 th) Tim 0.250 x 0-7 Sin 350 (*) E 3/288) 1350 (1) TW) so, N 1.258 0.250 x 0.7 sin N - 0.12546- 0.125/ h) N . The magnitude of the total force on the winte is 0.125 N