Read the following passage and answer the accompanying questions - Leaving Cert Physics - Question 11 - 2017

The definition of the ampere is based on the electromagnetic force between two parallel conductors. When current flows through two wires placed 1 meter apart in a vacuum, they experience a force of 2 x 10^-7 N per meter of length. This force arises from the interaction of the magnetic fields generated by the currents in the wires, which shows the principle that a current-carrying conductor experiences a force in a magnetic field.

The diagram should include four resistors arranged in a diamond shape, with two resistors on each side of the bridge. A galvanometer should be connected across the bridge to show the detection of current. An additional part should indicate the connection to a power source.

Telstar operated in a Medium Earth Orbit, which limits its coverage due to its orbiting path. Consequently, it could only facilitate communication during specific windows when it was above the horizon, leading to gaps in transmission where the satellite was not in geostationary orbit and thus not always above the same place on Earth.

Using the formula for orbital radius: $T^2 = \frac{4\pi^2 r^3}{GM}$ Substituting known values: $T = 2.6 \times 60 \times 60,$ $M = 6.0 \times 10^{24} \, kg$ We find: $r = 9.6 \times 10^6 \, m$

Light is transmitted through optical fibres by the principle of total internal reflection. A diagram should illustrate:

• The optical fibre structure showing the core and cladding.
• An incident light ray approaching the core-cladding interface at an angle greater than the critical angle, resulting in total internal reflection. The core must have a higher refractive index than the cladding to ensure that the light remains confined within the core.

To find the refractive index $n$, we can use the formula: $v = \frac{c}{n}$ where:

• $v$: speed of light in glass
• $c$: speed of light in a vacuum ($3 \times 10^8 \, m/s$) Given the distance and time: $d = 5500 \times 10^3 \, m$ $t = 29 \times 10^{-3} \, s$ Calculate $v = \frac{d}{t}$. Hence, $n = \frac{c}{v} = 1.58$

(i) in telegraph cables: electrons, (ii) in satellite signals: photons, (iii) in optical fibres: photons.