A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in Fig. 9.1.

Question 14

A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in Fig. 9.1.

Fig. 9.1 (not to scale)

Some of the corners of the slice are labelled.

A current I of 3.8 A is normal to face RSXY of the slice.

In aluminium, the number of free electrons per unit volume is 6.0 × 1028 m−3.

A uniform magnetic field of magnetic flux density B equal to 0.13 T is normal to face QRYZ of the aluminium slice in the direction from Q to P.

A Hall voltage VH is developed across the slice and is given by the expression

VH = BI / ntq

.

(a) Use Fig. 9.1 to state the magnitude of the distance t. [1]

(b) Calculate the magnitude of the Hall voltage VH. [2]

[Total: 3]

Reference: Past Exam Paper – June 2016 Paper 41 & 43 Q9

Solution:

(a)

0.10 mm

{In the equation, t is the thickness of materials through which the current passes.}

(b)

{V_H = BI / ntq

Thickness t should be in metres.

Charge of electrons: q = 1.60×10^-19}

VH = (0.13 × 3.8) / (6.0×10²⁸ × 0.10×10^-3 × 1.60×10^-19)

VH = 5.1 × 10^-7 V                             

An isolated spherical planet has a diameter of 6.8 × 106 m. Its mass of 6.4 × 1023 kg may be assumed to be a point mass at the centre of the planet.

Question 10

An isolated spherical planet has a diameter of 6.8 × 10⁶ m. Its mass of 6.4 × 10²³ kg may be assumed to be a point mass at the centre of the planet.

(a) Show that the gravitational field strength at the surface of the planet is 3.7 N kg^-1. [2]

(b) A stone of mass 2.4 kg is raised from the surface of the planet through a vertical height of 1800 m.

Use the value of field strength given in (a) to determine the change in gravitational potential energy of the stone.

Explain your working. [3]

(c) A rock, initially at rest at infinity, moves towards the planet. At point P, its height above the surface of the planet is 3.5 D, where D is the diameter of the planet, as shown in Fig. 1.1.

Fig. 1.1

Calculate the speed of the rock at point P, assuming that the change in gravitational potential energy is all transferred to kinetic energy. [4]

Reference: Past Exam Paper – November 2014 Paper 41 & 42 Q1

Solution:

(a)

Gravitational field strength, g = GM / R²

Gravitational field strength, g = (6.67×10^–11) × (6.4×10²³) / (3.4×10⁶)²

Gravitational field strength, g = 3.7 N kg^-1

(b)

Change in gravitational potential energy: ΔE_P = mgΔh

because (the vertical height is much smaller than the radius) Δh ≪ R (or 1800 m ≪ 3.4 × 10⁶ m), the gravitational field strength g is constant

{ΔE_P = mgΔh}

ΔE_P = 2.4 × 3.7 × 1800

ΔE_P = 1.6 × 10⁴ J

(c)

{Gravitational potential ϕ = - GM / x

Gravitational potential energy = mϕ = - GMm / x}

Gravitational potential energy = (–) GMm / x

{x is the separation between the rock and the CENTRE of the planet (since the planet is considered to be a point mass, all its mass is concentrated at the centre).}

{The GPE is all transferred to kinetic energy. So, considering the magnitudes,

½ mv² = GMm / x

giving v² = 2GM / x.}

v² = 2GM / x

{The rock is at a distance 3.5D from the surface of the planet. The surface is itself at a distance of D/2 = 0.5D from the centre of the planet.

So, separation x = 3.5D + 0.5D = 4D}

Separation x = 4D = 4 × 6.8×10⁶ m

{v² = 2GM / 4D}

v² = 2 × (6.67×10^–11) × (6.4×10²³) / (4 × 6.8×10⁶)

v² = 3.14 × 10⁶

Speed v = 1.8 × 10³ m s^–1