## A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The

Question

A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The transmitted radio wave has a frequency of 1.3 MHz. The magnetic field of the wave is parallel to the normal of the coil and has a maximum value of 1.7 x 10-13 T. Using Faraday’s Law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna in this time.

in progress 0
1 year 2021-08-31T14:50:59+00:00 1 Answers 5 views 0

The average  emf induce is   $$V = 2.625 * 10^{-5} \ V$$

Explanation:

From the question we are told that

The radius of the coil is  $$r = 0.30 \ m$$

The number of turns is  $$N = 420 \ turns$$

The frequency of the transition radio wave is  $$f = 1.3\ MHz = 1.3 *10^{6} Hz$$

The magnetic field is  $$B_,{max} = 1.7 * 10^{-13} \ T$$

The time taken for the magnetic field to go from zero to maximum is $$\Delta T = \frac{T}{4}$$

The period of the transmitted radio wave is  $$T = \frac{1}{f}$$

So

$$\Delta T = \frac{T}{4} = \frac{1}{4 f}$$

The potential difference can be mathematically represented as

$$V = NA (\frac{\Delta B}{\Delta T} )$$

$$V = NA ([B_{max} – B_{min} ] * 4f)$$

Where  $$B_{min} = 0T$$

substituting values

$$V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})$$

$$V = 2.625 * 10^{-5} \ V$$