Question

A triangle loop of wire shown in figure carries a current l = 3.0 A in the direction shown. The loop is in a uniform magnetic field that has magnitude B = 2.0 T with the same direction as the current in side PQ of the loop. 9 pts) a) Find the magnitude and direction of the magnetic force on each side of the triangle. (5 pts) b) What is the net magnetic force on the loop. (6 pts) c) If the loop is pivoted about an axis that lies along side PR, find the magnitude of net torque on the loop. 0.600 ml 0.800 m

Please solve this problem immidiately Thank you

Answer 1

PO = 0.6 m PR = 0.8 m QR = 10.6² 0.6² +0.8² I A 2 I 0.6m = 1m I tam o = B P 0.8m - tam (0) I=3 A B = 2T I 36.87 Force on due I el current carrying straight conducter magnetic field Current F = I l x B - in the direction of current flow = I l B Sin (&) Buy magnetic field For at angle between Fo= I l B sin (0) magnetic field and current flow ń whit vector along For PR (@xB) E = I&B sin (90°) I l B = 3x 0.882 4.8N (Into the plane) For Ro For IIB sin (9070) = 3x 1x2 x 8in (12687) Fra = 4.8N cout of the plane) - Fro=0 PR = FOR

Net toree prt - Ottis down word) = 0 + 4.8m Cinto the plane) + 4.8N Cont of the plane) = Net foree O the center q mass AB " the Look at figure again QR lies at A. perpendicular distance q PR. AB = VAR? -BR² - 10.5-0.42 from the line =0.3 m Torque = Fd, F Force Perpendicular distance da FPR is => d=0 acting atong shu on the axis opt itsef => Torque due to FIR=0 Torque due to For , Z = (FOR CAB) = 4.8 X 0.3 Hm - Net Torque Z net = 1.44 Nm