# A rectangular coil (1 = 24 cm, w = 19 cm, N = 27 turns) is placed in a magnetic field of 250 mT...

## Question:

A rectangular coil (1 = 24 cm, w = 19 cm, N = 27 turns) is placed in a magnetic field of 250 mT perpendicular to the plane of the coil. In how much time must the coil be withdrawn entirely from the B-field in order to reduce an average emf of 27.0 V across the coil?

In physics, mathematically faraday's law described as the electromotive force in a coil is equal to the product of the number of turns and the differential change of magnetic flux with time.

## Answer and Explanation:

• The length of rectangular coil is: {eq}L = 24\;{\rm{cm}} = 0.24\;{\rm{m}} {/eq}
• The width of rectangular coil is: {eq}w = 19\;{\rm{cm}} = 0.19\;{\rm{m}} {/eq}
• The number of turns is: {eq}N = 27 {/eq}
• The magnetic field is: {eq}B = 250\;{\rm{mT = 250}} \times {10^{ - 3}}{\rm{T}} {/eq}
• The emf across the coil is: {eq}\varepsilon = 27\;{\rm{V}} {/eq}

The area of rectangular coil is calculated as,

\begin{align*} A &= L \times w\\ &= 0.24\;{\rm{m}} \times 0.19\;{\rm{m}}\\ &= 0.0456\;{{\rm{m}}^2} \end{align*}

Recall the expression for the faraday?s Law.

$$\color{red}{\varepsilon = - N\dfrac{{d\phi }}{{dt}}}$$

Where, {eq}\phi {/eq} is the magnetic flux, which is equal to {eq}\phi = B \times A. {/eq}

The above expression can be written as,

\begin{align*} \varepsilon &= N\dfrac{{d\left( {BA} \right)}}{{dt}}\\ \varepsilon \int\limits_0^t {dt} &= NA\int\limits_0^B {dB} \end{align*}

Integrate the above equation.

\begin{align*} \varepsilon t &= NAB\\ t& = \dfrac{{NAB}}{\varepsilon } \end{align*}

Substitute the known values in above equation.

\begin{align*} t &= \dfrac{{27 \times 0.0456 \times 250 \times {{10}^{ - 3}}}}{{27}}\\ &= 0.0114\;{\rm{s}} \end{align*}

Thus, the time is {eq}\boxed{\color{blue}{0.0114\;{\rm{s}}}} {/eq}.