Question

It is a kind of electromagnetic balance in the figure as shown below. A metal bar connected to a constant current I is hanging on the right side. The area in the dashed box has a constant B-field, B = 800 T. Left arm has a length L1 = 2 m and right arm has a length L2 = 3 m. The horizontal length of the coil is 0.1 m. The total weight on the left side is m = 3,030 kg and the total weight on the right side is 20 kg. If the whole system is in balance, what is the current I on the right side? The positive current direction is defined as rightward. (g= 10 m/s2) Li BR | 0.1m

Answer 1

Please check the question

Either the weight on left side should be 3.030kg or 3030kg

I am attaching both possible solutions

Solution 1

Mass on left side is 3.030 kg

i A Given data: hq=2 m -L2= 3m length of the coil (1) = 0.2m weight on left side (m) = 3·030 kg. weight on right side (M) = 20Ky 1.1m B * Magnetic field (B) = 800 T - ( Into the paper accelelation due to gravity qe 10 mst Now magnetic force acting on wire (fm) = ILXB = IIB sino Here g=90° = IIB . . Direction of force perpendicular to wire A 18 FRD Now E Moment@_b_20 " mg 4 + FB L2 - Mg L2 zo

2.030x10x 2 + I X0.1 x 8003 20x10x3 31030 xlox2 + I (0-3 X 800] x3 - 20910 X3 20 I = 20x1003 - 3.030 X 1002 0:3X800 x3 - I = 0.7494)

Solution 2 , mass on left side is 3030 kg

Güven data: L, -2m 123m length of horizontal coil cl) = 0.1m mass on mass on left side (m, ) = 3030 kq right side (M) = 2014 strength of Magnetic field = 80oT ® |aceelvedtion "dece oto gravity C = 10mg-2 Assumption: Direction of current i toward Leight __ I P FE Now, magnetic force Acting on the coil (FB) FR = Il XB = Il B sino O=90° FR = IIB direction as Magnetic force perpendicular to coil in upward Wireetion. A L Now, FBD of Balance Now, E Moment @ B = 0 mgh, FB L₂ - Mgt2 = 0 e R 0 ma mg

3030x10x2 + IX 0:1 X800X3 - 20X10X3 = 0 I = 2081013 - 3030 X1082 0.18800x3 I tos7 - ? Pt? Here negative sign shows, direction should be toward left side of current 0