# A circuit is in a uniform magnetic field, that is into the page, and is decreasing in magnitude,...

## Question:

A circuit is in a uniform magnetic field, that is into the page, and is decreasing in magnitude, at the rate of 150 tesla/second. What current does the ammeter read?

## Induced Current:

The ammeter is the device to measure current flowing in the electrical circuit. A galvanometer can be made into an ammeter by connecting a shunt resistance in parallel to it. The rate of change of magnetic flux produces induced current.

## Answer and Explanation:

Lets us assume the following variables,

{eq}V \rightarrow \textrm{voltage across the terminals}\\ \epsilon \rightarrow \textrm{electromotive force}\\ \dfrac{d\phi}{dt} \rightarrow \textrm{rate of change of magnetic flux}\\ B \rightarrow \textrm{magnetic field} r \rightarrow \textrm{internal resistance in the circuit}\\ A \rightarrow \textrm{area of cross-section}\\ {/eq}

We know that the e.m.f (electromotive force) is the rate of change of magnetic flux which can be mathematically represented as,

{eq}\epsilon = \dfrac{d\phi}{dt} \Rightarrow \dfrac{d(A.B)}{dt}\\ \epsilon = A.\; \dfrac{d\phi}{dt}\Rightarrow (0.01\;\rm m^2)\dfrac{dB}{dt}\\ \epsilon = (0.01\;\rm m^2)(150\;\rm T/s)\\ \boxed{\epsilon \approx 1.5\;\rm V}......(1) {/eq}

The relation between the e.m.f and voltage (electric potential difference) is mathematically given as,

{eq}\epsilon = V + Ir\\ I = \dfrac{\epsilon - V}{r}\\ I = \dfrac{1.5\;\rm V - 1.0\;\rm V}{0.2\;\rm \Omega}\\ \boxed{I = 2.5\;\rm A} {/eq}

Following assumptions were made,

{eq}V = 1.0\;\rm V\\ r = 0.2\;\rm \Omega\\ A = 0.01\;\rm m^2 {/eq}