Question

A 107-turn circular coil of radius 2.41 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. during 0.115 s the magnetic field strength increases from 52.1 mt to 91.7 mt. find the magnitude of the average emf, in millivolts, that is induced in the coil during this time interval.

Answer 1

Answer:

Average emf in the coil: 0.0672 V.

Explanation:

Convert all units to standard SI units.

• Radius of the coil: $r=2.41\;\text{cm} = 2.41\times 10^{-2}\;\text{m}$.
• Initial magnetic field strength: $B = 52.1\;\text{mT} = 52.1\times 10^{-3}\;\text{T}$.
• Final magnetic field strength: $B = 91.7\;\text{mT} = 91.7\times 10^{-3}\;\text{T}$.

Consider Faraday's Law of Induction:

$\displaystyle \begin{aligned}\epsilon &= \text{Rate of change in}\;(N\cdot \phi)\\&=\text{Rate of change in}\; (N \cdot (B\cdot A\cdot \cos{\theta}))\end{aligned}$

where

• $N\cdot \phi$ is the magnetic flux linkage through the coil.
• $N$ is the number of turns in the coil.
• $\phi = B\cdot A\cdot \cos{\theta}$ is the magnetic flux through the coil.
• $B$ is the strength of the magnetic field,
• $A$ is the area of the coil,
• $\theta$ is the angle between the normal of the coil and the magnetic field. The coil is perpendicular to the magnetic field. As a result, the normal of the coil is parallel with the field. $\theta = 0$. $\cos{\theta} = 1$. $B\cdot A\cdot \cos{\theta} = B\cdot A$.

The coil is circular with a radius of $2.41\times 10^{-2}\;\text{m}$. As a result,

$A = \pi\cdot r^{2} = \pi\times (2.41\times 10^{-2})^{2} = 1.82467\times 10^{-3}\;\text{m}^{2}$.

Neither $N$ nor $A$ changes in this 0.115 seconds. As a result, the average rate of change in $N\cdot B\cdot A$ is the same as $N\cdot A$ times the average rate of change in $B$.

$\displaystyle \begin{aligned}\text{Average}\;\epsilon &= \text{Average Rate of Change in}\; (N\cdot (B\cdot A\cdot \cos{\theta}))\\&=\text{Average Rate of Change in}\; (N\cdot B\cdot A) \\&= (N\cdot A)\cdot \text{Average Rate of Change in}\;B\\&= 107\times 1.82467\times 10^{-3}\times \frac{91.7\times 10^{-3}- 52.1\times 10^{-3}}{0.115}\\ &=0.0672\;\text{V}\end{aligned}$.

All numbers in the question come with three sig. fig. Keep more sig. fig. than that in the calculation but round the final answer to three sig. fig.