2022_GCE_H3PH_9814_P1_SS [HCI]

2022 H3 A Level Suggested Solutions 1 (a) (i) The net torque generated on rod due to its weight is given by, 𝜏 = ∫ 𝑔 (𝑀 𝐿 ) 𝑑𝑙. 𝑥 = ∫ 𝑔 (𝑀 𝐿 ) 𝑑𝑙. (𝑙𝑠𝑖𝑛𝜃) = 𝑔𝑀𝑠𝑖𝑛𝜃 𝐿 (𝐿2 2 ) = 𝑀𝑔𝐿 𝑠𝑖𝑛𝜃 2 (Alternatively, since the rod is uniform, you may obtain the torque as follows: The torque due to the weight of the rod about A would be sin2 LMg   . As A= I , and 222 12 2 3 A ML L MLM= + =  I (The term 2 12 ML is the moment of inertia about centre of mass of the rod) We have, 2sin 23 3 sin 2 MgL ML g L    = = Comments: Several students did not consider the factor of two in their answers. (ii) Comparing with the given form, 1.5k = . Mg sin2 L 

(b) As the total energy is conserved, the gravitational PE is converted to the rotational kinetic energy of the rod, 2 21cos 0 02 2 3 3 cos L MLMg g L    + = +   = Comments: Using the conservation of energy is the most straightforward method. Some students erroneously used SUVAT equations, however, the angular acceleration was not constant. 2 (a) ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 Where ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ is the net magnetic flux along the loop, 𝜇0 is the magnetic permeability of free space, 𝐼𝑒𝑛𝑐 is the enclosed current and 𝑑ℓ is the small length on the circular path (loop). (b) (i) Since 𝐼1 = 𝐼2, ∫ 𝐽1 𝑑𝐴 = ∫ 𝐽2 𝑑𝐴 ∫ 𝐽1 (2𝜋𝑟 𝑑𝑟) 𝑟1 0 = ∫ 𝐽2 (2𝜋𝑟 𝑑𝑟) 𝑟3 𝑟2 𝜋𝑟1 2𝐽1 = 𝜋(𝑟3 2 − 𝑟2 2)𝐽2 𝐽2 = 𝑟1 2 𝑟3 2 − 𝑟2 2 𝐽1 (ii)1 Draw a circular Amperian loop of radius 𝑟1 centred on the axis of the wire, ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 = 𝜇0𝐼1,𝑒𝑛𝑐 = 𝜇0(𝜋𝑟1 2𝐽1) 𝐵(2𝜋𝑟1) = 𝜇0𝜋𝑟1 2𝐽1 𝐵 = 1 2 𝜇0𝐽1𝑟1 (ii)2 Draw a circular Amperian loop of radius 𝑟2 centred on the axis of the wire, ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 = 𝜇0𝐼1 = 𝜇0(𝜋𝑟1 2𝐽1) 𝐵(2𝜋𝑟2) = 𝜇0𝜋𝑟1 2𝐽1 𝐵 = 𝜇0𝐽1𝑟1 2 2𝑟2

(iii) The shapes of the graph are: • 0 to r1: linear • (Maximum at r1) • r1 to r2: curve • r2 to r3: curve • r3 to r4: zero Calculations for each region (no need to show): Repeating the working in 2(b)(ii)1 and 2 but considering a loop with a general radius 𝑟, we have: 𝐵 = { 1 2 𝜇0𝐽1𝑟, 0 < 𝑟 ≤ 𝑟1 𝜇0𝐽1𝑟1 2 2𝑟 , 𝑟1 ≤ 𝑟 ≤ 𝑟2 That is, 𝐵 ∝ 𝑟 for 0 < 𝑟 ≤ 𝑟1 and 𝐵 ∝ 1 𝑟 for 𝑟1 ≤ 𝑟 ≤ 𝑟2 For 𝑟2 ≤ 𝑟 ≤ 𝑟3: Draw a circular Amperian loop of radius 𝑟 centred on the axis of the wire, where 𝑟2 ≤ 𝑟 ≤ 𝑟3. (Note that the current in the metal braid flows in the opposite direction, so the net current is 𝐼1 − 𝐼2,𝑒𝑛𝑐.) ∮ 𝐵⃗ ∙ 𝑑ℓ⃗ = 𝜇0𝐼𝑒𝑛𝑐 = 𝜇0(𝐼1 − 𝐼2,𝑒𝑛𝑐) = 𝜇0(𝜋𝑟1 2𝐽1 − 𝜋(𝑟2 − 𝑟2 2)𝐽2) 𝐵(2𝜋𝑟) = 𝜇0 (𝜋𝑟1 2𝐽1 − 𝜋(𝑟2 − 𝑟2 2) 𝑟1 2 𝑟3 2 − 𝑟2 2 𝐽1) 𝐵(2𝜋𝑟) = 𝜇0𝜋𝐽1𝑟1 2 (1 − 𝑟2 𝑟3 2 − 𝑟2 2 + 𝑟2 2 𝑟3 2 − 𝑟2 2) 𝐵(2𝜋𝑟) = 𝜇0𝜋𝐽1𝑟1 2 𝑟3 2 − 𝑟2 2 (𝑟3 2 − 𝑟2) 𝐵 = 𝜇0𝐽1𝑟1 2 2(𝑟3 2 − 𝑟2 2) (𝑟3 2 𝑟 − 𝑟) If you expand the bracket you will get two terms: r1 r2 r3 r4 conductor insulator plastic jacket metal braid B x 0 0 Bmax

• The first term is proportional to 1/𝑟, and initially this dominates. However it gets smaller as 𝑟 increases. • The second term