All categories

3 years ago

The magnetic field a distance 2 cm from a long straight current-carrying wire is 2 × 10–5 t. the current in the wire is:

Physics

1 answer:

Sliva [168]3 years ago

6 0

Current in the wire = 2 A

Explanation:

the magnetic field is given by

B= \frac{\mu i}{2\pi r}

μo= 4π x 10⁻⁷ Tm/A

i= current

r=0.02 m

B = magnetic field= 2 x 10⁻⁵ T

2 x 10⁻⁵= (4π x 10⁻⁷)(i) / (2π*0.02)

i=2 A

You might be interested in

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig

r-ruslan [8.4K]

Answer:

2406 miles

Explanation:

Let A be the starting position, B the junction position and C the final position after flying the 3.5 hrs. Also, let b be the distance from the starting point:

$\angle ABC =(180-10) \textdegree=170\textdegree$

#Distance traveled in 1.5hrs is;

$c=690x1.5\\=1035mi$

#Distance traveled in next two hrs:

$a=690\times 2\\=1380mi$

#Now using the Cosine Rule:

$b^2=a^2+c^2-2ab\cos B\\\\=1380^2+1035^2-2(1380)(1035)cos170\textdegree\\\\b^2=5788.83\\\\b\approx 2406.00 \ mi$

Hence, the pilot is 2406 miles from her starting position.

4 0

3 years ago

A proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

lawyer [7]

The change in potential energy of the proton is 5.6 x $10^{-17}$ Joule

<h3>What is a Uniform Electric Field ?</h3>

A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.

Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

The distance d = 10 cm = 0.1 m

Electric field E = 3.5 KN/C

Proton charge q = 1.6 x $10^{-19}$ C

The Work done = Fd

but F = Eq

Recall that Electric field E = F/q = V/d

Where V = potential difference.

Let us first calculate the V

E = V/d

V = Ed

Substitute all the parameters into the formula above

V = 3.5 × 10³ × 0.1

V = 350 v

from F/q = V/d

make F the subject of formula and substitute it in work formula

F = Vq/d

W.D = Vq/d x d

W.D = Vq

Substitute all the parameters into the formula above

W.D = 350 x 1.6 x $10^{-19}$

W.D = 5.6 x $10^{-17}$ J

Work done = Energy = Potential Energy

Therefore, the change in potential energy of the proton is 5.6 x $10^{-17}$ <em> Joule</em>

<em />

Learn more about Electric Field here: brainly.com/question/14372859

#SPJ1

7 0

2 years ago

Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i

densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0

3 years ago

a 2,000-kilogram railroad car moving at 8m/s to the right collides with a 6,000-kilogram railroad car moving at 3m/s to the west

astra-53 [7]

A freight car of mass 20,000 kg moves along a frictionless level railroad track ... After the push the skateboarder II moves with a velocity of 2 m/s to ... After the collision the cars stick to each other and ... diver jumps with a velocity of 3 m/s in opposite ... A 10 kg object moves at a constant velocity 2 m/s to the right and collides

3 0

3 years ago

A rock with a mass of 8 kg falls straight down from a height of 7 m. What work is done?

fenix001 [56]

F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps

7 0

3 years ago