Title 16 Electromagnetism Tutorial Soln.pdf ✅

St. Andrew’s Junior College H2 Physics 16-4 4(a)(i) Magnetic force provides for the centripetal force. Bqvy = mvy 2 r (3.0 × 10-5 )(1.6 × 10-19)(6.7 × 106 × sin 40°) = (9.11 × 10-31)(6.7 × 106 × sin 40°)2 / r r = 0.817 m (ii) vy = r(2π / T) 6.7 × 106 × sin 40° = (0.81737)(2π / T) T = 1.19 × 10-6 s (iii) Using sx = uxt, p = (6.7 × 106 × cos 40°)(1.19249 × 10-6 ) p = 6.12 m (b) With the parallel plate, a uniform E-field will set up such that there is now a horizontal electrical force acting on the electron towards the right. Hence, the electron will now accelerate horizontally, resulting in increasing horizontal component of its velocity. With T being constant, the pitch p will increase. 5(a) B due to WX = μ0 I 2πd = (4π × 10−7 )(9.0) 2π(50 × 10−3) = 0.000036 T, out of the page B due to YZ = μ0 I 2πd = (4π × 10−7 )(12.0) 2π(50 × 10−3) = 0.000048 T, vertically upward Resultant B = √0.0000362 + 0.0000482 = 6.0 × 10-5 T Direction = tan-1 ( 0.000048 0.000036) = 53.1° above the horizontal (b) Since P is still 50 mm away from YZ, B due to YZ remains at 0.000048 T. To get resultant B = 0 at point P, B due to WX must be equal to 0.000048 as well. Hence, B due to WX = μ0 I 2πd 0.000048 = (4π × 10−7 )(9) 2π(d) d = 37.5 mm Therefore, distance of WX away from YZ = 50 + 37.5 = 87.5 mm away from YZ orientation of WX should be parallel to YZ direction of current to be in the same direction as current in YZ