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Pachacha [2.7K]
3 years ago
11

The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t

he wire?
Physics
1 answer:
Tanzania [10]3 years ago
8 0

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

B = \frac{\mu_oI}{2\pi r}\\\\r =  \frac{\mu_oI}{2\pi B}

where,

r = distance of point P from wire = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = 2 A

B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T

Therefore,

r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\

<u>r = 0.05 m = 5 cm</u>

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Two long, straight wires are parallel and are separated by a distance of d = 0.210 m. The top wire in the sketch carries current
love history [14]

Answer:

1.88\cdot 10^{-5} T, inside the plane

Explanation:

We need to calculate the magnitude and direction of the magnetic field produced by each wire first, using the formula

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I is the current

r is the distance from the wire

For the top wire,

I = 4.00 A

r = d/2 = 0.105 m (since we are evaluating the field half-way between the two wires)

so

B_1 = \frac{(4\pi\cdot 10^{-7})(4.00)}{2\pi(0.105)}=7.6\cdot 10^{-6}T

And using the right-hand rule (thumb in the same direction as the current (to the right), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is inside the plane

For the bottom wire,

I = 5.90 A

r = 0.105 m

so

B_2 = \frac{(4\pi\cdot 10^{-7})(5.90)}{2\pi(0.105)}=1.12\cdot 10^{-5}T

And using the right-hand rule (thumb in the same direction as the current (to the left), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is also inside the plane

So both field add together at point P, and the magnitude of the resultant field is:

B=B_1+B_2 = 7.6\cdot 10^{-6} T+1.12\cdot 10^{-5}T=1.88\cdot 10^{-5} T

And the direction is inside the plane.

3 0
3 years ago
A force F moves a load from the bottom of a slope to the top
marysya [2.9K]

Answer:

i think A

Explanation:

3 0
3 years ago
A triangle ∆P QR has vertices P(3, 2, −4), Q(1, 0, −4), R(2, 1, 1). Use the distance formula to decide which one of the followin
777dan777 [17]

Answer:

a. FALSE

b.TRUE

C. FALSE

Explanation:

The formula fot the distance between two points is given as

d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2} +(z_{2}-z_{1})^{2}}\\

hence we determine the distances between all the points

a.P(3,2,-4), Q(1,0,-4), R(2,1,1)

PQ=\sqrt{(1-3)^{2} +(0-2)^{2} +(-4-(-4))^{2}}\\PQ=\sqrt{4+4+0}\\ PQ=\sqrt{8}

For point PR

we have

PR=\sqrt{(2-3)^{2} +(1-2)^{2} +(1-(-4))^{2}}\\PR=\sqrt{1+1+9}\\ PR=\sqrt{11}\\

|PQ|\neq |PR|

B. For point RP

RP=\sqrt{(3-2)^{2} +(2-1)^{2} +(-4-1)^{2}}\\RP=\sqrt{1+1+25}\\ RP=\sqrt{27}

for point RQ  we have

RQ=\sqrt{(1-2)^{2} +(0-1)^{2} +(-4-1)^{2}}\\RQ=\sqrt{1+1+25}\\ RQ=\sqrt{27}

|RP|=|R Q|

C.

QP=\sqrt{(3-1)^{2} +(2-0)^{2} +(-4+4)^{2}}\\QP=\sqrt{4+4+0}\\ QP=\sqrt{8}

For point Q R

QR=\sqrt{(2-1)^{2} +(1-0)^{2} +(1-(-4))^{2}}\\QR=\sqrt{1+1+9}\\ QR=\sqrt{11}\\

QP\neq QR

6 0
3 years ago
a runner goes from 5 m/s to a stop in 10 seconds. Calculate his acceleration. Show your work and include units. ​
AnnyKZ [126]
Hope this helps you out

5 0
3 years ago
If a projectile is fired straight up at a speed of 10 m/s, the time it takes to reach the top of its path is about
UNO [17]

Answer:

A. The time it takes the projectile to reach the top of its path is about 1 second.

Explanation:

Hi there!

The equation of the velocity of a projectile fired straight up is the following:

v = v0 + g · t

Where:

v = velocity of the projectile.

v0 = initial velocity.

g = acceleration due to gravity (≅ -9.8 m/s² considering the upward direction as positive)

t = time.

When the projectile reaches the top of its path, its velocity is zero, then, using the equation of velocity, we can solve it for the time:

v = v0 + g · t

0 = 10 m/s - 9.8 m/s² · t

t = -10 m/s / -9.8 m/s²

t = 1.0 s

The time it takes the projectile to reach the top of its path is about 1 second.

7 0
3 years ago
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