Question

A 25-turn circular coil of wire has diameter 1.00 m. It is placed with its axis along the direction of the Earth’s magnetic field of 50.0 μT and then in 0.200 s is flipped 180°. An average emf of what magnitude is generated in the coil?

Answer 1

Answer:

The magnitude of average emf is 9.813 mV.

Explanation:

Given that,

Number of turns = 25 turns

Diameter = 1.00 m

Magnetic field = 50.0μm

Time = 0.200 s

Angle = 180°

We need to calculate the area of the coil

Using formula of area

$A=\pr r^2$

$A=\pi\times(\dfrac{1.00}{2})^2$

$A=0.785\ m^2$

We need to calculate the average emf

Using formula of emf

$E=NAB\cos\theta$

$E=NA(\dfrac{B\cos0^{\circ}-B\cos180^{\circ}}{dt}$

$E=\dfrac{2NA}{dt}$

Where, N = number of turns

A = Area

B = magnetic field

Put the value into the formula

$E=\dfrac{2\times25\times0.785\times50.0\times10^{-6}}{0.200}$

$E=9.813\ mV$

Hence, The magnitude of average emf is 9.813 mV.

Answer 2

The average emf generated will be equal to EMF =9.813mv

What will be the average emf?

It is given that

Number if the turns N=25

Diameter =1m

Earths magnetic field B=50$\muT$$\muT$

Time =0.2sec

Angle =180 Degree

The area of the coil is

$A= \pi r^2$

$A=\pi 0.5^2$

$A=0.758m^2$

Now for calculating the EMF

$E=NABcos\theta$

$E=NA(\dfrac{Bcos(0)-Bcos(180)}{dt}$

$E=\dfrac{2NA}{dt}$

$\dfrac{2\times25\times0.785\times50\times10^{-6}}{0.2}$

$E=93813 Mv$

Thus the average emf generated will be equal to EMF =9.813mv

To know more about EMF follow

https://brainly.com/question/13744192