Ask question

Ask question

All categories

2 years ago

11

A 170.4-turn circular loop of radius 0.08 meters with a magnetic field B passing through the area bounded by the loop at right a

ngles has an Emf of 0.050 Volts induced around the loop by the rising magnetic field. How fast is the magnetic field rising in tesla per second

1 answer:

jolli1 [7]2 years ago

8 0

Answer:

The speed of the magnetic field rising is $\frac{\Delta B}{\Delta t}$ $=0.0147 \ T/s$

Explanation:

Mathematically Induced Emf is as follows

$e = - \frac{\Delta B}{\Delta t} *N * Area$

The objective of this solution is to obtain speed in Tesla per second which the same as $\frac{\Delta B}{\Delta t}$ since the unit of magnetic field B is Tesla and Time is seconds

So making $\frac{\Delta B}{\Delta t}$ the subject

$\frac{\Delta B}{\Delta t } =\frac{Emf}{N*Area}$

$The \ Area = \pi r^2$

$= 3.142 * (0.08)^2$

$=0.020m$

N is given as 170.4 turns

$Therefore \ \frac{\Delta B}{\Delta t} = \frac{0.050}{170.4*0.02}$

$=0.0147 \ T/s$

You might be interested in

Practice Exercises Name: : Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below a) If Billy-Joe kicks

babymother [125]

Answer:

a) The answer is 11,7m

b) The time it takes to fall will be shorter

Explanation:

We will use the next semi-parabolic movement equations

$H=Hi+Viy*t+1/2*g*t$

$X=Xi+Vx*t$

Where g(gravity acceleration)=9,81m/s^2

Also Xi, Hi and Viy are zero, as the stones Billy-Jones is kicking stay still before he moves them, so we take that point as the reference point

The first we must do is to find how much time the stones take to fall, this way:

$t=(5.40m)/(3.50m/s)$

Then t=1,54s

After that we need to replace t to find H, this way

$H=(1/2)*(9,81m/s^2)*(1,54s)^2$

Then H=11,7m

b) The stones will fall faster as the stones will be kicked harder, it will cause the stones move faster, it means, more horizontal velocity. In order to see it better we could assume the actual velocity is two times more than it is, so it will give us half of the time, this way:

$t=(5,40m)/(2*3,50m/s)$

Then, t=0,77

4 0

2 years ago

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0

3 years ago

What happens to a day of light that slows down when it hits a new medium at an angle

Anika [276]

Light travels as transverse waves and faster than sound. It can be reflected, refracted and dispersed. Ray diagrams show what happens to light in mirrors and lenses. Eyes and cameras detect light.

4 0

3 years ago

Seja uma carga Q=1.2×10(-8)C no vácuo, distando 40 cm de um ponto M e 50 cm de um ponto N . Determine os pontencias no ponto M e

Len [333]

<span>Americano, por favor?</span>

4 0

3 years ago

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

$q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }$

6 0

2 years ago