# A 100-turn, 2.00-cm diameter coil is at rest in a horizontal plane. A uniform magnetic field 60...

## Question:

A 100-turn, 2.00-cm diameter coil is at rest in a horizontal plane. A uniform magnetic field 60 degrees away from vertical increases from 0.5 T to 1.50 T in 0.6 seconds. What is the induced emf in the coil?

## Electromagnetic Induction:

When a conducting loop is placed in a magnetic field an emf is induced in the loop until there is a change in the magnetic flux through the loop. This emf is called induced emf and the current thus produced is called induced current.

The magnetic flux of a field {eq}B {/eq} through an area {eq}A {/eq} is given by:

{eq}\begin{align*} \Phi_B &= \int \vec B \cdot d\vec A \\ &= BA & \text{[when the magnetic field is perpendicular to the cross-section of the area]} \end{align*} {/eq}

Faraday's law gives the induced emf {eq}\xi {/eq} in a conducting loop in a magnetic field as the time rate of change of the magnetic flux {eq}\Phi_B {/eq} through the loop:

{eq}\displaystyle \begin{align*} \xi &= -\frac{d\Phi_B}{dt} \\ &= -\frac{\Delta \Phi_B}{\Delta t} & \text{[when the flux changes steadily over time]} \end{align*} {/eq}

The minus sign tells us that the induced emf will be directed such that it opposes the change in flux.

## Answer and Explanation:

**Given:**

- Number of turns in a solenoid is {eq}N = 100 \ \text {turns} {/eq}

- The radius of a tube is {eq}r = \dfrac {d} {2} = \dfrac {2.00} {2} = 1.00 \ cm = 1.00 \times 10^{-2} \ m {/eq}

- The change in magnetic field is {eq}dB = B_2 - B_1 = 1.50 - 0.5 = 1.00 \ T {/eq}

- The angle between the magnetic field lines and the normal (perpendicular) to A is {eq}\theta = 60^\circ {/eq}

- The time duration is {eq}dt = 0.6 \ s {/eq}

**Let**

- The average induced emf in the coil is {eq}\varepsilon {/eq}.

After plugging the given values in the formula of average induced emf, we get the magnitude of induced emf in the coil.

The value of the magnitude of induced emf is:

{eq}\begin{align} \varepsilon & = - N \dfrac { d\phi} {dt} \\ & = - N \dfrac { d } {dt} (BA\cos\theta ) \\ & = - N A \cos\theta \dfrac { dB } {dt} \\ & = - N A \cos\theta \dfrac { (B_2 -B_1 ) } {dt} \\ & = - (100) \pi (1.00 \times 10^{-2})^2 \ cos60^\circ \left \{ \dfrac { 1.00 } { 0.6 } \right \} \\ & = -( 1.57\times 10^{-2}) ( 1.67 ) \\ \implies \varepsilon & = -26.2 \ mV \\ \end{align} {/eq}

The induced EMF opposes the increasing current (Lenz's Law) which means that emf is negative.

**Hence, the average induced emf in the coil is {eq}26.2 \ mV {/eq}.**

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from High School Physics: Help and Review

Chapter 13 / Lesson 10