Question

A closely wound rectangular coil of 80 turns has dimensions 25.0 cm by 40.0 cm. The plane of the coil is rotated from a position in which it makes an angle of 37.0 degrees with a magnetic field of 1.10 Τ to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf ε induced in the coil?

What is the magnitude of the average emf E induced as the coil is rotated?

E= ________ V

Answer 1

Concepts and reason

The concepts used to solve this problem are the rate of change of magnetic flux and Faraday’s law of electromagnetic induction.

The emf is generated when a coil of wire is moved into a magnetic field. The induced current will create a magnetic field to oppose the magnetic field in the coil.

First, Calculate the flux difference and then by using the Faraday law of induction, the magnitude of induced emf can be calculated.

Fundamentals

The expression for the induced emf developed in the coil is,

$\varepsilon = - N\left( {\frac{{\Delta \phi }}{{\Delta t}}} \right)$

Here, $\varepsilon$ is the emf induced in the coil, N is the number of turns, $\Delta \phi$ is the change in magnetic flux, and $\Delta t$ is the change in time.

Express the relation between magnetic flux in terms of angle.

$\phi = BA\cos \theta$

Here, A is the area of the coil and $\phi$is the angle between the direction of magnetic field and normal to the plane of the coil, and B is the magnetic field.

Express the area of the rectangular coil.

$A = lb$

Here, $l$is the length of the coil and $b$ is the breadth of the coil.

Substitute $25.0{\rm{ cm}}$ for $l$ and $40.0{\rm{ cm}}$for b to find A.

$\begin{array}{c}\\A = \left( {25\,{\rm{cm}} \times \frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1\,{\rm{cm}}}}} \right)\left( {40\,{\rm{cm}} \times \frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1\,{\rm{cm}}}}} \right)\\\\ = 0.1\,{{\rm{m}}^2}\\\end{array}$

Since the initial flux produced in the coil is,

${\phi _{{\rm{initial}}}} = BA\cos \left( {90^\circ - \theta } \right)$

Here, $\cos \left( {90^\circ - \theta } \right)$ indicates the angle made between the plane and the magnetic field.

Substitute $37^\circ$ for $\theta$, $0.1\,{{\rm{m}}^2}$ for $A$, and $1.10\,{\rm{T}}$ for $B$ to find ${\phi _{{\rm{initial}}}}$.

$\begin{array}{c}\\{\phi _{{\rm{initial}}}} = \left( {1.10\,{\rm{T}}} \right)\left( {0.1\,{{\rm{m}}^2}} \right)\cos \left( {90^\circ - 37^\circ } \right)\\\\ = 0.066\,{\rm{T}} \cdot {{\rm{m}}^2}\\\end{array}$

Since the final flux produced in the coil is,

${\phi _{{\rm{final}}}} = BA\cos \theta$

Substitute $0^\circ$ for $\theta$, $0.1\,{{\rm{m}}^2}$ for $A$, and $1.10\,{\rm{T}}$ $B$ to find ${\phi _{{\rm{final}}}}$.

$\begin{array}{c}\\{\phi _{{\rm{final}}}} = \left( {1.10\,{\rm{T}}} \right)\left( {0.1\,{{\rm{m}}^2}} \right)\\\\ = 0.11\,{\rm{T}}{\rm{.}}{{\rm{m}}^2}\\\end{array}$

The average emf induced in the coil is,

$\varepsilon = - N\left( {\frac{{\Delta \phi }}{{\Delta t}}} \right)$

Rewrite the expression in terms of initial and final magnetic flux.

$\varepsilon = - N\left( {\frac{{{\phi _{{\rm{final}}}} - {\phi _{{\rm{initial}}}}}}{{\Delta t}}} \right)$

Substitute 80 turns for $N$, $0.0600\,{\rm{s}}$ for $\Delta t$, $0.11\,{\rm{T}}{\rm{.}}{{\rm{m}}^2}$ for ${\phi _{{\rm{final}}}}$, and $0.066\,{\rm{T}} \cdot {{\rm{m}}^2}$ for ${\phi _{{\rm{initial}}}}$ in the above expression to find the magnitude of the emf.

$\begin{array}{c}\\\varepsilon = - 80\left( {\frac{{0.11\,{\rm{T}}{{\rm{m}}^2} - 0.066\,{\rm{T}}{{\rm{m}}^2}}}{{0.06\,{\rm{s}}}}} \right)\\\\ = - 58.67\,{\rm{V}}\\\\\left| \varepsilon \right| = 58.67\,{\rm{V}}\\\end{array}$

Ans:

The magnitude of the average emf induced in the coil is$58.67\,{\rm{V}}$.