Ask question

Ask question

All categories

11111nata11111 [884]

3 years ago

13

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15

cm from the wire, what is the maximum current the wire can carry?

1 answer:

dem82 [27]3 years ago

6 0

Answer:

37.725 A

Explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as

$B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}$

Inserting the values

$0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}$

i = 37.725 A

You might be interested in

Which weather condition commonly occurs along a cold front

aleksklad [387]

Precipitation is your answer. according to my science textbook, in a cold front, warm air and cold air interact and cause precipitation.

7 0

3 years ago

Read 2 more answers

Nina [5.8K]

Number 18 is C, i thin

5 0

3 years ago

Read 2 more answers

Two soccer players kick a soccer ball back and forth along a straight line. The first player kicks the ball 16 m to the right to

Oliga [24]

Answer:

a

The total distance is $d_t =19.1 m$

b

The displacement is

$D_t = 12.9m$

Explanation:

From the question we are told that

Distance traveled by the ball for first player $d_f = 16m$ to the right

Distance traveled by the ball for second player $d_s = 3.1 m$ to the left

The total distance traveled by the ball is mathematically represented as

$d_t = d_f + d_s$

Substituting values

$d_t = 16 + 3.1$

$d_t =19.1 m$

The displacement is mathematically represented as

$D_t = d_f -d_s$

This is because displacement deal with direction and from the question we are told that right is positive and left is negative

Substituting values

$D_t = 16 -3.1$

$D_t = 12.9m$

5 0

3 years ago

All of the following are factors affecting flow rate except what?

faust18 [17]

Answer:

c. Vessel side holes

Explanation:

The "Poiseuille formula" which is given by $\\\begin{aligned} \small Q& = \small \frac{\pi r^4}{8 \eta}.\frac{\Delta P}{\Delta L}\\\end{aligned}$ describes the volumetric flow rate ( $\small Q$ ) through tubular sections.
Here, $\Delta P,\,\, \Delta L,\,\, r,\,\, \eta$ represent the injection pressure difference, the length of the section, the radius of the section and the viscosity index of the fluid that flows through the section respectively.
With this, one can confirm that all the factors except the vessel side holes affect the flow rate.
Side holes, however, are a factor that could give a measure of how much volume would flow to a particular location. In such a situation the flow rate remains unchanged and one location would receive a lower volume (not the whole) as some volume would spill out at the side holes.

#SPJ4

5 0

2 years ago

A 10.0-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wi

Veronika [31]

A) 1.05 N

The power dissipated in the circuit can be written as the product between the pulling force and the speed of the wire:

$P=Fv$

where

P = 4.20 W is the power

F is the magnitude of the pulling force

v = 4.0 m/s is the speed of the wire

Solving the equation for F, we find

$F=\frac{P}{v}=\frac{4.20 W}{4.0 m/s}=1.05 N$

B) 3.03 T

The electromotive force induced in the circuit is:

$\epsilon=BvL$ (1)

where

B is the strength of the magnetic field

v = 4.0 m/s is the speed of the wire

L = 10.0 cm = 0.10 m is the length of the wire

We also know that the power dissipated is

$P=\frac{\epsilon^2}{R}$ (2)

where

$R=0.350 \Omega$ is the resistance of the wire

Subsituting (1) into (2), we get

$P=\frac{B^2 v^2 L^2}{R}$

And solving it for B, we find the strength of the magnetic field:

$B=\frac{\sqrt{PR}}{vL}=\frac{\sqrt{(4.20 W)(0.350 \Omega)}}{(4.0 m/s)(0.10 m)}=3.03 T$

5 0

3 years ago