Physics, asked by pramodvns11, 2 months ago

State and explain two laws which are used to find the direction of deflection in magnetic

campass needle kept near a current carrying conductor.​

Answers

Answered by poojasri0132
1

Answer:

Explanation:

How is this described?

There are two key laws that describe electromagnetic induction:

Faraday's law, due to 19ᵗʰ century physicist Michael Faraday. This relates the rate of change of magnetic flux through a loop to the magnitude of the electro-motive force \mathcal{E}EE induced in the loop. The relationship is

\mathcal{E} = \frac{\mathrm{d}\Phi}{\mathrm{d}t}E=

dt

E, equals, start fraction, d, \Phi, divided by, d, t, end fraction

The electromotive force or EMF refers to the potential difference across the unloaded loop (i.e. when the resistance in the circuit is high). In practice it is often sufficient to think of EMF as voltage since both voltage and EMF are measured using the same unit, the volt. [Explain]

Lenz's law is a consequence of conservation of energy applied to electromagnetic induction. It was formulated by Heinrich Lenz in 1833. While Faraday's law tells us the magnitude of the EMF produced, Lenz's law tells us the direction that current will flow. It states that the direction is always such that it will oppose the change in flux which produced it. This means that any magnetic field produced by an induced current will be in the opposite direction to the change in the original field.

Lenz's law is typically incorporated into Faraday's law with a minus sign, the inclusion of which allows the same coordinate system to be used for both the flux and EMF. The result is sometimes called the Faraday-Lenz law,

\mathcal{E} = -\frac{\mathrm{d}\Phi}{\mathrm{d}t}E=−

dt

E, equals, minus, start fraction, d, \Phi, divided by, d, t, end fraction

In practice we often deal with magnetic induction in multiple coils of wire each of which contribute the same EMF. For this reason an additional term NNN representing the number of turns is often included, i.e.

\mathcal{E} = -N \frac{\mathrm{d}\Phi}{\mathrm{d}t}E=−N

dt

Similar questions