Question

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.5 T is directed along the positive z-direction, which is upward. Of the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wire loop during the extraction process.

Answer 1

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Step-by-step explanation:

The average emf induced can be calculated from the formula

`Emf = -N(\Delta \phi)/(\Delta t)`

Where
`N` is the number of turns

`\Delta \phi` is the change in magnetic flux

`\Delta t` is the time interval

The change in magnetic flux is given by

`\Delta \phi = \phi _(f) - \phi _(i)`

Where
`\phi _(f)` is the final magnetic flux

and
`\phi _(i)` is the initial magnetic flux

Magnetic flux is given by the formula

`\phi = BAcos(\theta)`

Where
`B` is the magnetic field

`A` is the area

and
`\theta` is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is,
`\theta = 0^(o)`

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

`A = \pi r^(2)`

∴
`A = \pi (0.15)^(2)`

`A = 0.0225\pi`

∴
`\phi_(i) = (1.5)(0.0225\pi)cos(0^(o) )`

`\phi_(i) = (1.5)(0.0225\pi)`

( NOTE:
`cos (0^(o)) = 1` )

`\phi_(i) = 0.03375\pi` Wb

For
`\phi_(f)`

The field pointed upwards, that is
`\theta = 90^(o)`. Since
`cos (90^(o)) = 0`

Then

`\phi_(f) = 0`

Hence,

`\Delta \phi = 0- 0.03375\pi`

`\Delta \phi = - 0.03375\pi`

From the question

`\Delta t = 2.8 ms = 2.8 * 10^(-3) s`

Here,
`N` = 1

Hence,

`Emf = -N(\Delta \phi)/(\Delta t)` becomes

`Emf = -(1)(-0.03375\pi)/(2.8 * 10^(-3) )`

`Emf = 37.9 V`

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.