2019 GCE A Level 9814 H3 P1 SS [HCI]

Hwa Chong Institution (College) 1 2019 A level paper (9814) suggested solutions 1 (a) Draw the elliptical shape. Correct labelling of the maximum distance between the centre of ellipse to the point on the orbit. 2 (a) Gauss’s Law for magnetic fields states that the total magnetic flux through a closed surface is zero, 0B dA = (b) A magnetic field line that does not form an open loop will imply the presence of magnetic monopoles at the ends of the field line. As magnetic monopoles do not exist, magnetic field lines will always form a closed loop. (c) If a magnetic monopole exists, the magnetic flux calculated based on a closed surface that enclose the magnetic monopole will be non -zero which is inconsistent with Gauss’s Law for magnetic fields. 3 (a) 0B ds = I Ampere’s Law states that where B is the magnetic field strength, ds is a line element along the integration path, is the permeability of free space and I is the current passing through the area enclosed by the integration path. (b) (i) Apply Ampere’s Law along a circular path of radius d centered at the axis of the wire that encloses the wire, when d is larger than the radius of the wire, ( ) 0 0 0 2 2 B ds Bd B d     = = =  I I I (ii) When d is shorter than the radius of the wire, assuming that the current is uniform across the cross section of the wire, Sun Planet semi-major axis, a

Hwa Chong Institution (College) 2 ( ) 0 2 0 2 0 2 2 2 B ds dBd r Bd r      = =   =  I I I Hence, we have the following graph (iii) A magnetic field exists inside the wire as the current flows through all parts of the cross section of the wire, hence when we apply Ampere’s Law over a circular path inside the wire, the path will enclose a fraction of the current that flows in the wire resulting in a non-zero magnetic field in the wire. 4 (a) The centre of mass must lie somewhere along the line of symmetry (blue dashed line). It should also lie to the left of the green dashed line. COMMENT: The C.M. of the completed square is exactly at r. The C.M. of the missing square is exactly at s, on the right of r. So the C.M. of the L-shape must be left of r. X r s

Hwa Chong Institution (College) 3 (b) (i) cosx R t = siny R t =− R is the distance between A and X. (ii) COMMENT: Sketch should show increasing “pitch”. (iii) sin cos yy R t v R t  =−  =− Maximum upward velocity due to rotation 1 (0.50)(25) 12.5 m s R − = = = ()v u at=+ Velocity of centre of mass (9.81)t= Solve for 12.5 9.81 1.274 s t t = = X t=0 t=T t=2T t=3T t=4T

Hwa Chong Institution (College) 4 (iv) 22 0.251325T  === s 1.274 s 1.274 0.2513 5.07tT= =  = After the 5th rotation, the vertical velocity is always downward. So the trajectory stops intersecting itself. The last intersection occurs during the 4th rotation. From the sketch, it looks like the