Help pleaseee is this correct?

A long, straight wire lies along the z-axis and carries a 4.20-A current in the +z-direction. Find the magnetic field (magnitude

lana [24]

Complete question:

A long, straight wire lies along the z-axis and carries a 4.00-A current in the +z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire centered at the origin:

(a) x = 2.00 m, y = 0, z = 0;

(b) x = 0, y = 2.00 m, z = 0;

(c) x = 2.00 m, y = 2.00 m, z = 0; (d) x = 0, y = 0, z = 2.00 m.

Answer:

a) $5.25*10^-1^1 T$ ;

b) $5.25*10^-^1^1 T$ ;

c) $1.86*10^-^1^1 T$ ;

d) 0

Explanation:

Given

I = 4.20A

P = 0.500mm

a) r= (2.00m) i

∆ $T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2$ ;

$B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (2.00)^3$ ;

$(B= 5.25*10^-^1^1T) j$

b) r = (2.00m) j

∆ $T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2$ ;

$B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (2.00)^3$ ;

$(B= 5.25*10^-^1^1T) i$

c) r = (2.00m) (i+j)

$(i/j) r = \sqrt{2} (2.00m)$ ;

∆ $T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2$ ;

$B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (\sqrt{2})(2.00)^3$ ;

$(B= 1.86*10^-^1^1T)(i-j)$

d) r = (2.00m) k

∆ $T*r = (0.500*10^-^3)(2.00) k*k = 0$ ;

B = 0

8 0

3 years ago

A ball is thrown up with a vertical velocity of 15 m/s and a horizontal velocity of 50 m/s. Calculate how many seconds it will t

DerKrebs [107]

Answer:

It will take 23 hours and 29 minutes

Explanation:

5 0

2 years ago

harge of uniform density (100 nC/m3) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical sur

Elden [556K]

Answer:

The magnitude of the electric field is 8.47 N/C

Explanation:

Given;

uniform charge density, λ = 100 nC/m³

inner radii of the cylinder, r = 1.0 mm and 3.0 mm

distance from the symmetry axis, R = 2.0 mm

$Volume =\pi (R^2 -r^2)l\\\\Volume =\pi ((2*10^{-3})^2 -(1*10^{-3})^2)l\\\\Volume =\pi (4*10^{-6} - 1*10^{-6})l\\\\Volume = 3*10^{-6} \pi l \ m^3$

Area = 2πrl

Area =2π(2 x 10⁻)l

Volume = A x d

d = Volume / Area

$d = \frac{V}{A} = \frac{3*10^{-6}*\pi*l}{4\pi *10^{-3} l} = 75 *10^{-5} \ m$

the magnitude of the electric field

$E = \frac{\lambda *d}{\epsilon_o} = \frac{100*10^{-9} *75*10^{-5}}{8.854*10^{-12}} \\\\E = 8.47 \ N/C$

Therefore, the magnitude of the electric field is 8.47 N/C

6 0

3 years ago

When measuring the diameter of a cylinder the following measurements are obtained.

Arlecino [84]

Answer:

a. 1.80 %

b. 0.60 %

c. 1.20 %

d. 0.60 %

e. 0.60 %

Explanation:

In order to calculate percent uncertainties in each case, we forst need to calculate the average value:

Average Diameter = (16.4 mm + 16.8 mm + 16.9 mm + 16.6 mm +16.8 mm)/5

Average Diameter = 16.7 mm

Now, the formula for percent uncertainty is:

% Uncertainty = (Uncertainty/Average) * 100 %

where,

Uncertainty = |Value - Average|

For Each Case:

a. 16.4 mm:

Uncertainty = |16.4 mm - 16.7 mm| = 0.3 mm

Therefore,

% Uncertainty = (0.3 mm/16.7 mm) * 100%

% Uncertainty = 1.80 %

b. 16.8 mm:

Uncertainty = |16.8 mm - 16.7 mm| = 0.1 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

c. 16.4 mm:

Uncertainty = |16.9 mm - 16.7 mm| = 0.2 mm

Therefore,

% Uncertainty = (0.2 mm/16.7 mm) * 100%

% Uncertainty = 1.20 %

d. 16.6 mm:

Uncertainty = |16.6 mm - 16.7 mm| = 0.1 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

e. 16.8 mm:

Uncertainty = |16.8 mm - 16.7 mm| = 0.3 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

% Uncertainty = 0.60 %

7 0

3 years ago

1. What do you notice from the picture? 2. Did anyone get the correct answer?

Fudgin [204]

Answer:

1. I notice nothing because is no picture it probably didnt work

2. idk

6 0

3 years ago

Help pleaseee is this correct?​

Help pleaseee is this correct?