Question

To monitor the breathing of a hospital patient, a thin belt is girded around the patient's chest as in the figure below. The belt is a 195 turn coil. When the patient inhales, the area encircled by the coil increases by 42.0 cm². The magnitude of earth's magnetic field is 50.0 µT and makes an angle of 28.0° with the plane of the coil. Assuming a patient takes 1.60 s to inhale, find the magnitude of the average induced emf in the coil during that time.
|| =
µV?

Answer 1

Concepts and reason

The concepts that are to be used to solve the given problem are induced emf and magnetic flux, and magnetic field.

Express the magnetic flux equation and then express the induced emf as rate of change of magnetic flux. Then, replace magnetic flux expression in to induced emf equation to obtain the expression for induced emf in terms of given quantities. Finally, substitute the values into the derived equation for induced emf to find the induced emf.

Fundamentals

The magnetic flux $\Phi$ is,

$\Phi = BA\cos \theta$

Here, $B$ is the strength of magnetic field, $A$ is the area of the loop, and $\theta$ is the angle between the magnetic field and the normal of the plane of the loop.

The average induced emf is,

$\varepsilon = - N\frac{{\Delta \Phi }}{{\Delta t}}$

Here, N is the number of turns, and $\frac{{\Delta \Phi }}{{\Delta t}}$ is the rate of change of flux.

$\Phi = BA\cos \theta$

The average induced emf is,

$\varepsilon = - N\frac{{\Delta \Phi }}{{\Delta t}}$

Replace $\Phi$ with $BA\cos \theta$ .

$\begin{array}{l}\\\varepsilon = - N\frac{{\Delta \left( {BA\cos \theta } \right)}}{{\Delta t}}\\\\\varepsilon = - NB\cos \theta \left( {\frac{{\Delta A}}{{\Delta t}}} \right)\\\end{array}$

The magnitude of the average induced emf is,

$\left| \varepsilon \right| = NB\cos \theta \left( {\frac{{\Delta A}}{{\Delta t}}} \right)$

Substitute 195 for N, ${62.0^{\rm{o}}}$ for $\theta$ , 1.60 s for $\Delta t$ , $50.0{\rm{ }}\mu {\rm{T}}$ for B, and $42.0{\rm{ c}}{{\rm{m}}^2}$ for $\Delta A$ in $\left| \varepsilon \right| = NB\cos \theta \left( {\frac{{\Delta A}}{{\Delta t}}} \right)$ .

$\begin{array}{c}\\\left| \varepsilon \right| = \left( {195} \right)\left( {50.0{\rm{ }}\mu {\rm{T}}\left( {\frac{{{{10}^{ - 6}}{\rm{ T}}}}{{1{\rm{ }}\mu {\rm{T}}}}} \right)} \right)\cos {62.0^{\rm{o}}}\left( {\frac{{42.0{\rm{ c}}{{\rm{m}}^2}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}{{1.60{\rm{ s}}}}} \right)\\\\ = 12 \times {10^{ - 6}}{\rm{ V}}\left( {\frac{{1{\rm{ }}\mu {\rm{V}}}}{{{{10}^{ - 6}}{\rm{ V}}}}} \right)\\\\ = 12{\rm{ }}\mu {\rm{V}}\\\end{array}$

Ans:

The magnitude of average induced emf is $12{\rm{ }}\mu {\rm{V}}$ .