Ask question

Ask question

All categories

Zigmanuir [339]

2 years ago

12

What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in ai

r? Give your answer in units of tesla.

1 answer:

olga nikolaevna [1]2 years ago

3 0

Answer:

The magnetic flux density is $2.11\times10^{-6}\ T$

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

$B =\dfrac{\mu_{0}I}{2r}$

Where,

r = radius

I = current

Put the value into the formula

$B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}$

$B=2.11\times10^{-6}\ T$

Hence, The magnetic flux density is $2.11\times10^{-6}\ T$

You might be interested in

Please help! i'm horrible at this

frez [133]

Answer:

a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²

Explanation:

You may or may not need the negative sign, depending on how the question designer was thinking about the problem.

4 0

2 years ago

A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe

Sindrei [870]

Answer:

The magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ .

Explanation:

Given that,

Charge, $q=-8.5\ \mu C=-8.5\times 10^{-6}\ C$

Speed of the charged particle, $v=9\times 10^6\ m/s$

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, $F=5.9\times 10^{-3}\ N$

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

$F=qvB\ \sin\theta$

B is the magnetic field.

$B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T$

So, the magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ . Hence, this is the required solution.

4 0

2 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

2 years ago

Identify the region in the image where lighter elements, such as hydrogen or helium, are more likely to be found in a liquid or

Annette [7]

Lower in an arrangement

3 0

2 years ago

Read 2 more answers

The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0

AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t represented by function :

$p(t)=t^{2} +2t-4$

Now,

When function p(t) = 11 , t will be

$p(t)=t^{2} +2t-4$

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0

t = 3 , -5

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = $\frac{position}{time}$

therefore to get the value of velocity from function p(t)

we have to differentiate the function with respect to time

$\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)$

v(t) = 2t + 2

where v(t) = velocity at that time

as t = 3 for p(t)=11

so ,

v(t) = 2t + 2

v(t) = 2*3 + 2

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

To know more about function here:

brainly.com/question/12431044

#SPJ4

4 0

1 year ago