The average speed of a plane was 600 km/hr. How long did it take

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago

A complete path through which charge can flow is an electric what?

grigory [225]

<span>A complete path through which charge can flow is an "Electric Circuit"

Hope this helps!</span>

5 0

3 years ago

Read 2 more answers

How many moles are in 73.4 grams of Phosphorus?

mylen [45]

23.2

hope this helped

4 0

3 years ago

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

$\theta$ = Angle of slide = 20.7°

$\mu$ = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

$W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J$

The work done by the force of gravity is 23.52092 J

8 0

2 years ago

On the Moon the acceleration due to gravity is about one sixth that on Earth. If a golfer on the Moon imparted the same initial

swat32

Explanation:

We know that that the range of the ball on the earth

$R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}$

therefore, range of the ball on moon

$R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}$

$R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}$

therefore,

$R_{moon}=6R_{earth}$

Therefore, the range of ball will be 6 times on the moon than that on earth

6 0

2 years ago