Question

A 90-turn square coil of side 20.0 cm rotates about a vertical axis at ω = 1.65 103 rev/min as indicated in the figure below. The horizontal component of Earth's magnetic field at the coil's location is equal to 2.00 10-5 T.


(a) Calculate the maximum emf induced in the coil by this field.
in mV

(b) What is the orientation of the coil with respect to the magnetic field when the maximum emf occurs?
a.The plane of the coil is parallel to the magnetic field?
b.The plane of the coil is oriented 45° with respect to the magnetic field?
c.The plane of the coil is perpendicular to the magnetic field?

Answer 1

Concepts and reason

The concepts used to solve this problem are emf generated in a rotating coil and Faraday’s law.

Initially, use the relation between area and length of the side of a square to find to find the area of cross section.

Then, use the relation between magnetic field intensity, area, number of turns of the coil, and rate of rotation of the coil to find the maximum emf induced in the coil by the field.

Finally, use the concept of maximum value of emf induced in the coil to find the orientation of the coil with respect to the magnetic field when the maximum emf occurs.

Fundamentals

“The Faraday’s law of induction states that whenever there is a change in the in the magnetic flux enclosed by the circuit, an emf will induce in the circuit and this emf is proportional to the negative of rate of change of the magnetic flux.”

Expression for the emf generated in the rotating coil is,

$\varepsilon = BAN\omega \sin \left( \theta \right)$

Here, the emf generated in the rotating coil is $\varepsilon$ , magnetic field intensity is $B$ , area is $A$ , number of turns in the coil is $N$ , rate of rotation of the coil is $\omega$ , and the angle between the normal to the coil is $\theta$ .

Expression for the area of the square is,

$A = {a^2}$

Here, the area of the square is $A$ and length of the side of a square is $a$ .

(a)

Expression for the area of the square is,

$A = {a^2}$

Substitute $20.0\,{\rm{cm}}$ for $a$ .

$\begin{array}{c}\\A = {\left( {\left( {20.0\,{\rm{cm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{{{10}^2}\,{\rm{cm}}}}} \right)} \right)^2}\\\\ = 0.04\,{{\rm{m}}^2}\\\end{array}$

The maximum value of emf is when the coil is in the plane of the field.

Expression for the maximum value of emf generated in the rotating coil is,

$\varepsilon = BAN\omega$

Substitute $2.00 \times {10^{ - 5}}\,{\rm{T}}$ for $B$ , $0.04\,{{\rm{m}}^2}$ for $A$ , $90\,{\rm{turns}}$ for $N$ , and $1.65 \times {10^3}\,{\rm{rev/min}}$ .

$\begin{array}{c}\\\varepsilon = \left( {2.00 \times {{10}^{ - 5}}\,{\rm{T}}} \right)\left( {0.04\,{{\rm{m}}^2}} \right)\left( {90\,{\rm{turns}}} \right)\left[ {\left( {1.65 \times {{10}^3}\,\frac{{{\rm{rev}}}}{{{\rm{min}}}}} \right)\left( {\frac{{2\left( {3.14\,{\rm{rad}}} \right)\left( {1\,{\rm{min}}} \right)}}{{60\,{\rm{s}}}}} \right)} \right]\\\\ = 0.0124\,{\rm{V}}\\\\{\rm{ = }}\left( {0.0124\,{\rm{V}}} \right)\left( {\frac{{{{10}^3}\,{\rm{mV}}}}{{1\,{\rm{V}}}}} \right)\\\\ = 12.4\,{\rm{mV}}\\\end{array}$

(b)

One of the incorrect option is,

(b)The plane of the coil is oriented $45\,^\circ$ with respect to the magnetic field.

Expression for the emf generated in the rotating coil is,

$\varepsilon = BAN\omega \sin \left( \theta \right)$

At $45^\circ$ , the value of emf is,

$\begin{array}{c}\\\varepsilon = BAN\omega \sin 45^\circ \\\\ = \frac{{BAN\omega }}{{\sqrt 2 }}\\\end{array}$

Another incorrect option is,

(c)The plane of the coil and magnetic field are perpendicular to each other.

When the plane of the coil and magnetic field are perpendicular to each other, the angle between the area vector and the magnetic field is $0^\circ$ .

Thus, the value of emf is,

$\begin{array}{c}\\\varepsilon = BAN\omega \sin 0^\circ \\\\ = 0\\\end{array}$

The correct option is,

a)The plane of the coil is parallel to the magnetic field.

When the plane of the coil and magnetic field are parallel to each other, the angle between the area vector and the magnetic field is $90^\circ$ .

Thus, the value of emf is,

$\begin{array}{c}\\\varepsilon = BAN\omega \sin 90^\circ \\\\ = BAN\omega \\\end{array}$

Ans: Part a

Thus, the maximum value of induced emf is ${\bf{12}}{\bf{.4}}\,{\bf{mV}}$ .

Part b

Thus, the plane of the coil and magnetic field are parallel to each other.