Ask question

Ask question

All categories

3 years ago

5

A 5-turn square loop (10 cm along aside, resistance = 4.0 ) is placed in a magnetic field that makes an angle of30o with the pla

ne of the loop. The magnitude of thisfield varies with time according to B =0.50t2, where t is measured in s andB in T. What is the induced current in the coil att = 4.0 s?
1. 25 mA
2. 50mA
3. 13mA
4. 43mA
5. 5.0 mA

1 answer:

andreev551 [17]3 years ago

4 0

Answer:

Explanation:

Area A of the coil = .1 x .1 = .01 m²

no of turns n = 5

magnetic field B = .5 t²

Flux Φ perpendicular to plane passing through it.= nBA sin30

rate of change of flux

dΦ/dt = nAdBsin30 / dt

= nA d/dt (.5t²x .5 )

= nA x 2 x .25 x t

At t = 4s

dΦ/dt = nA x 2

= 5x .01 x 2

= .1

current = induced emf / resistance

= .1 / 4

= .025 A

= 25 mA.

You might be interested in

The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

vovikov84 [41]

Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;

$\frac{F_oR_o^2}{Q_o*Q_o}=\frac{F_{o2}R_{o2}^2}{3Q_o*Q_o}.............(3)$

Therefore;

$\frac{F_oR_o^2}{Q_o^2}=\frac{F_{o2}R_{o2}^2}{3Q_o^2}.............(4)$

From equation (4) we now make the new force $F_{o2}$ the subject of formula as follows;

${F_oR_o^2}*{3Q_o^2}=F_{o2}R_{o2}^2*{Q_o^2}$

$Q_o$ then cancels out from both side of the equation, hence we obtain the following;

$3{F_oR_o^2}=F_{o2}R_{o2}^2.............(4)$

From equation (4) we can now write the following;

$F_{o2}=\frac{3F_oR_o^2}{R_{o2}^2}$

This could also be expressed as follows;

$F_{o2}=3F_o(\frac{R_o}{R_{o2}})^2$

3 0

4 years ago

A force acts upon an object causing displacement of the object is the definition of ?

chubhunter [2.5K]

The answer is displacement

3 0

3 years ago

klasskru [66]

The statement which describes how a machine can help make work easier is that It can put out more force than the input force by decreasing the distance over which force is applied, therefore the correct option is option A.

<h3>What is work done?</h3>

The total amount of energy transferred when a force is applied to move an object through some distance

The work done is the multiplication of applied force with displacement.

Work Done = Force * Displacement

As work done depends both on the force as well as the displacement force and be reduced by reducing the displacement if the same amount of work is performed by the machine.

The correct answer is option A since the statement that describes how a machine might assist in making work easier says that it can put out greater force than the input force by reducing the distance over which force is exerted.

Learn more about work done from here

brainly.com/question/13662169

#SPJ1

5 0

2 years ago

A ball is dropped from the top of a 77 m building. With what speed does the ball hit the ground? _________ m/s

vitfil [10]

Answer:

38.87 m/s

Explanation:

Given that the ball is dropped from a height = 77 m

u = 0 m/s

s = 77 m

a = g = 9.81 m/s²

Applying the expression as:

$v^2-u^2=2as$

Applying values as:

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 77+0^2}\\\Rightarrow v=38.87\ m/s$

<u>The speed with which the ball hit the ground = 38.87 m/s</u>

3 0

3 years ago

How much mass does the sun lose through nuclear fusion per second?

Mekhanik [1.2K]

Answer:

1.10^6 kg of mass per second

Explanation:

All energy lost by the sun comes from nuclear fusion.

Sun loses energy at 2.5*10^{19}J per hour, that is 9*10^{22}J/s

To find the mass lost by the sun in liberation of energy you use the famous Einstein's equation:

$E=mc^2\\\\m=\frac{E}{c^2}=\frac{9*10^{22}}{(3*10^{8}m/s)^2}=1*10^6kg$

hence, the sun liberates 1.10^6 kg of mass per second

3 0

4 years ago