# An 47-turn coil has square loops measuring 0.141 m along a side and a resistance of 5.77 Ω....

## Question:

An 47-turn coil has square loops measuring 0.141 m along a side and a resistance of 5.77 Ω. It is placed in a magnetic field that makes an angle of 31.1 {eq}^o {/eq} with the plane of each loop. The magnitude of this field varies with time according to B = 2.49 t{eq}^3 {/eq}, where t is measured in seconds and B in Tesla. What is the induced current in the coil at t = 1.35 s?

## Magnetic Field:

The magnitude of the magnetic field can be calculated using the number of turns of the coil, current in the coil, cross-sectional area of the coil, and the resistance of the coil. The magnetic field is a vector quantity as it has both magnitudes as well as direction.

## Answer and Explanation: 1

**Given Data**

- The number of turn is {eq}N = 47 {/eq}.

- The length of the coil is {eq}l = 0.141\;{\rm{m}} {/eq}.

- The resistance of the coil is {eq}R = 5.77\;\Omega {/eq}.

- The angle of the magnetic field is {eq}\theta = 31.1^\circ {/eq}.

- The magnitude of the magnetic field is {eq}B = 2.49{t^3} {/eq}.

- The time is {eq}t = 1.35\;{\rm{s}} {/eq}.

The expression for the magnetic flux passing through the coil is,

{eq}\begin{align*} \varphi & = NBA\cos \theta \\[0.3 cm] &= NB\left( {l \times l} \right)\cos \theta \end{align*} {/eq}

Substituting the given values in the above expression, we will get

{eq}\begin{align*} \varphi & = \left( {47} \right)\left( {2.49{t^3}} \right)\left( {0.141\;{\rm{m}} \times 0.141\;{\rm{m}}} \right)\cos 31.1^\circ \\[0.3 cm] &= 1.992{t^3} \end{align*} {/eq}

The expression for the induced emf in the coil is,

{eq}\begin{align*} \varepsilon & = \dfrac{{d\varphi }}{{dt}}\\[0.3 cm] &= \dfrac{d}{{dt}}\left( {1.992{t^3}} \right)\\[0.3 cm] &= 1.992\left( {3{t^2}} \right)\\[0.3 cm] &= 5.976{t^2} \end{align*} {/eq}

Substituting the given values in the above expression, we will get

{eq}\begin{align*} \varepsilon & = 5.895{\left( {1.35\;{\rm{s}}} \right)^2}\\[0.3 cm] &= 10.89126\;{\rm{V}}\\[0.3 cm] &\approx10.89\;{\rm{V}} \end{align*} {/eq}

The expression for the induced current in the coil is,

{eq}I = \dfrac{\varepsilon }{R} {/eq}

Substituting the given values in the above expression, we will get

{eq}\begin{align*} I &= \dfrac{{10.89\;{\rm{V}}}}{{5.77\;\Omega }}\\[0.3 cm] &= 1.88\;{\rm{A}} \end{align*} {/eq}

Thus, **the current induced in the coil is {eq}\boxed{1.88\;{\rm{A}}}
{/eq}.**

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Chapter 7 / Lesson 2Magnetic fields fill the space around all magnets, but they're impossible to detect with our own senses. We'll use a common tool to map out a magnetic field and then discuss ferromagnetic materials to see how a magnetic field can be used to create new magnets.