Question

A 120-turn, 9.604-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 12 degrees away from vertical increases from 0.316 T to 5.553 T in 98.254 s. Determine the emf induced in the coil.

Answer 1

Given:

The number of turns is,

`n=120`

The diameter of the coil is

`\begin{gathered} d=9.604\text{ cm} \\ =9.604*10^(-2)\text{ m} \end{gathered}`

The angle of the magnetic field with the coil is

`\theta=12\degree`

The change in a magnetic field is from

`0.316\text{ T to 5.553 T}`

in

`t=98.254\text{ s}`

To find:

The induced emf

Step-by-step explanation:

The induced emf in the coil is,

`\xi=-nA(dB)/(dt)cos\theta`

Here, the area of the coil is,

`\begin{gathered} A=\pi(d^2)/(4) \\ =\pi*((9.604*10^(-2))^2)/(4) \\ =7.244*10^(-3)\text{ m}^2 \end{gathered}`

The induced emf is,

`\begin{gathered} \xi=-120*7.244*10^(-3)*(5.553-0.316)/(98.254)cos12\degree \\ =-0.045\text{ V} \end{gathered}`

Hence, the induced emf is 0.045 V.