Question

A 1300-turn coil of wire that is 2.10 cm in diameter is in a magnetic field that drops from 0.130 T to 0 T in 12.0 ms. The axis of the coil is parallel to the field.

Question: What is the emf of the coil? (in V)

Please explain

Answer 1

Concepts and reason

The concepts required to solve the given questions emf of the coil.

Initially, convert the value of the radius from cm to m. Later, substitute the values of different components in the formula of emf of coil. Finally, calculate the emf of coil.

Fundamentals

The expression emf of coil is as follows:

$\varepsilon = - NA\frac{{\Delta B}}{{\Delta t}}$

Here, $\varepsilon$ is the emf of coil, N is the number of turns, A is the area of the coil, $\Delta B$ is the change in the magnetic field, and $\Delta t$ is the change in the time.

The relation between the radius and the diameter is as follows:

$R = \frac{D}{2}$

Here, R is the radius and D is the diameter.

The expression for the area of coil is as follows:

$A = \pi {R^2}$

Substitute 2.10 cm for D in the equation $R = \frac{D}{2}$.

$\begin{array}{c}\\R = \frac{{2.10{\rm{ cm}}}}{2}\\\\ = 1.05{\rm{ cm}}\\\\{\rm{ = }}1.05{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)\\\\ = 0.0105{\rm{ m}}\\\end{array}$

Substitute 3.14 for $\pi$ and 0.0105 m for R in the equation $A = \pi {R^2}$.

$\begin{array}{c}\\A = \left( {3.14} \right){\left( {0.0105{\rm{ m}}} \right)^2}\\\\ = 0.00034{\rm{ }}{{\rm{m}}^2}\\\end{array}$

Substitute 1300 for N, $0.00034{\rm{ }}{{\rm{m}}^2}$ for A, $- 0.130{\rm{ T}}$ for $\Delta B$, and $12.0{\rm{ ms}}$ for $\Delta t$ in the equation $\varepsilon = - NA\frac{{\Delta B}}{{\Delta t}}$.

$\begin{array}{c}\\\varepsilon = - \left( {1300} \right)\left( {0.00034{\rm{ }}{{\rm{m}}^2}} \right)\left( {\frac{{ - 0.130{\rm{ T}}}}{{12.0{\rm{ ms}}}}} \right)\\\\ = - \left( {1300} \right)\left( {0.00034{\rm{ }}{{\rm{m}}^2}} \right)\left( {\frac{{ - 0.130{\rm{ T}}}}{{12.0{\rm{ ms}}\left( {\frac{{{{10}^{ - 3}}{\rm{ s}}}}{{1.00{\rm{ ms}}}}} \right)}}} \right)\\\\ \approx 4.9{\rm{ V}}\\\end{array}$

Ans:

The emf of the coil is equal to 4.9 V.