Neutral Atom - Higher Physics - Solved Past Paper, Exams of Physics

Download Neutral Atom - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! Higher Physics 1A PHYS1131 2006-S1 Answers 1. 12 Marks Total (a) A neutral atom is one with no net charge. The number of electrons is the same as the number of protons. (b) A negatively charged atom has one or more excess electrons over the number of protons. (c) (i) Charge 1 +q at (0,a), Charge 2 +3q at (2a, a), Charge 3 -q at (0, -a). Force between charges q1 and q2 a distance r apart is given by ! F 12 = k q 1 q 2 r 2 directed towards each other if the charges have opposite signs, and away from each other if they have the same sign. Charges 1 and 2 are 2a apart. Thus ! F12 = k q 3q (2a) 2 = 0.75k q a " # $ % & ' 2 directed along the positive x-axis. Charges 1 and 3 are ! (2a) 2 + (2a) 2 = 8a apart and the line connecting them is at 45° to the axes. Thus ! F23 = k "q 3q ( 8a) 2 = "0.375k q a # $ % & ' ( 2 ; i.e. directed towards each other. Thus, taking components along the x- and y-axes: x-axis: ! Fx = k q a " # $ % & ' 2 0.75 ( 0.375cos(45°)[ ] = k q a " # $ % & ' 2 0.485 y-axis: ! Fy = k q a " # $ % & ' 2 (0.375sin(45°)[ ] = (k q a " # $ % & ' 2 0.265 Therefore, the net force on charge 3q is given by ! F = k q a " # $ % & ' 2 0.485i ( 0.265 j[ ]. (ii) Potential at the origin is given by ! V = k qi ri " where charge qi is at distance ri from the origin. Thus ! V = k q a " q a + 3q a 2 + (2a) 2 # $ % % & ' ( ( = k 3q 5a =1.34k q a . 2. 6 Marks Total (a) Applying Gauss’s Law, the surface must enclose a positive net total charge, since ! q " 0 =# > 0 . (b) (i) Only the charge inside the radius R contributes to the flux. i.e. applying Gauss’s law, ! " = q # 0 . (ii) For a sphere of radius 2a we must include the total charge from both the ring and the point charge. Charge on the ring is 2πaλ. Thus, Gauss’s law gives ! " = q + 2#a$ % 0 . 5 13 Marks Total (a) Applying the right hand rule (motion up, field out of page), the proton experiences a force directed from left to right across the page, and so it veers to the right. It will proceed to follow a circle in the clockwise direction, as it always experiences a force perpendicular to its direction of motion. If instead the particle were a negatively-charged electron the path would veer to the left and then continue to move in a circle in the anti-clockwise direction. At the same speed, the electron’s circle would have a much smaller radius. (this comes from ! mv 2 r = qvB, hence r = mv qB , but they don’t need to prove this) (b) For each segment we have I = 5.00 A and B = 0.020 T j. The force on a current carrying wire is given by F = I l x B, where l is the vector denoting the length and direction of the wire. Resolve l into components along each section of the wire. (i) For segment ab l = -0.40 m j. Hence F = 5.00 x -0.40 x 0.020 j x j N = 0 N. (ii) For segment bc l = +0.40 m k. Hence F = 5.00 x 0.40 x 0.020 k x j N = -0.040 i N. (iii) For segment cd l = -0.40 m i + 0.40 m j. Hence F = 5.00 x 0.40 x 0.020 (-i x j + j x j) N = -0.040 k N. (iv) For segment da l = +0.40 m i - 0.40 m k. Hence F = 5.00 x 0.40 x 0.020 (i x j - k x j) N = 0.040 (k + i) N. (c) Since F = q v x B, then the acceleration produced by a magnetic field on a charged particle of mass m and charge q is a = q/m (v x B). This is perpendicular to the direction of the motion. For the acceleration to change the speed, a component of the acceleration must be along the direction of the velocity. It is not, so that the constant magnetic force changes the direction of the particle, but not its speed. 6. 9 Marks Total (a) Ampere’s law states that ! B.ds = µ 0 I" where the integral is around a closed path and I is the total steady current passing though any surface bounded by the path. Gauss’s law states that ! E .dA = q " 0 # where the integral is over a closed surface and q is the charge enclosed by that surface. Both laws use the concept of “flux” – the “flow” of field lines through a surface to determine the field strength. They also relate the integral of the field over a closed geometrical figure to a fundamental constant multiplied by the source of the appropriate field. The geometrical figure is a surface for Gauss’s law and a line for Ampere’s law; Ampere is the 2D analogue of Gauss. (b) From Ampere’s law, the magnetic field at point a is given by ! B a = µ 0 I a 2"r a where Ia is the net current through the area of the circle of radius ra, which in this case is 1.00A out of the page. Hence ! B a = (4"10#7 Tm /A)(1.00 A) 2" (1.00 $10#3 m) = 2.00 10#4 = 200 µT towards the top of the page (direction from right hand rule). Similarly, at point b: ! B b = µ 0 I b 2"r b where Ib is the net current through the area of the circle of radius rb, which in this case is 1.00-3.00 A = -2.00 A into the page. Therefore ! B b = (4"10#7 Tm /A)(2.00 A) 2" (3.00 $10#3 m) =133 µT towards the bottom of the page. 7. 8 Marks Total (a) Faraday’s Law of Induction states that ! " = # d$ dt ; i.e. the emf, ε, generated in a circuit is proportional to minus the rate of change of magnetic flux, Φ, through that circuit. (b) ! " = # d$ dt with Φ=BA, and A the area of the flux linked. Thus ! " = #B dA dt where ! dA dt is the rate of change of area = av (as only side a is cutting new flux). So ! " = #Bav . Now ! " = IR for a current I, so that IR=Bav, taking the absolute value. In equilibrium (i.e. constant speed, v), the weight balances the opposing force. The force on a current carrying conductor is given F=BIa, and from Lenz’s law it must be upwards (so as to oppose the change). Thus ! mg = BIa with ! I = Bav /R (from above). Hence ! B 2 av R = mg or ! B = mgR a 2 v = 0.5 9.8 2 1 2 8 = 1.11T .