Ways that industry and agriculture use physical properties to separate substances

The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00μT. (a) At what distance is it 0.100μ

maw [93]

The distance should be 4m from the wire in order to get the magnetic field of 0.100μ .

The magnitude and direction of the magnetic field due to a straight wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by $B=\frac{u_0I}{2\pi r}$ ...(1)
Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.

It is given that magnetic field 40.0 cm away from a straight wire is 1.00μT having current 2.00 A .

From equation (1) magnetic field 40.0 cm = 0.4m away from a straight wire is 1.00μT which is given by $1.00=\frac{u_0I}{2\pi \times0.4}$ .....(2)

From equation (1) magnetic field 'r' m away from a straight wire is 0.100μT which is given by $0.100=\frac{u_0I}{2\pi \times r}$ ...(3)

On dividing equation (2) by (3) , we get

$\frac{1}{0.1} =\frac{r}{0.4} \\\\r=4m$

Learn more about magnetic field here :

brainly.com/question/27939568

#SPJ4

4 0

2 years ago

Which of these is NOT an example of a reference direction?

Vikentia [17]

Answer:

A

Explanation:

8 0

4 years ago

In one experiment, the teacher determined that the force on the wire was 2.14 mN

blsea [12.9K]

Answer:

in ine experiment the teacher determined

3 0

3 years ago

The moon is 3x10^5 km away from Nepal and the mass of the moon is 7x10^22 kg. Calculate the force with which the Moon pulls ever

hammer [34]

Answer:

Approximately $5.19 \times 10^{-5}\; \rm N$ .

Explanation:

Let $G$ denote the gravitational constant. ( $G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}$ .)

Let $M$ and $m$ denote the mass of two objects separated by $r$ .

By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:

$\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}$ .

In this question: $M = 7 \times 10^{22}\; \rm kg$ is the mass of the moon, while $m = 1\; \rm kg$ is the mass of the water. The two are $r = 3\times 10^{5}\; \rm km$ apart from one another.

Important: convert the unit of $r$ to standard units (meters, not kilometers) to reflect the unit of the gravitational constant $G$ .

$\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m$ .

$\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}$ .

5 0

3 years ago

A solid disc with a radius of 5.00 m and a mass of 20.0 kg is initially at rests and lies on the plane of the paper. A smaller s

drek231 [11]

Answer:

Explanation:

This problem is based on conservation of angular momentum.

moment of inertia of larger disc I₁ = 1/2 m r² , m is mass and r is radius of disc . I

I₁ = .5 x 20 x 5²

= 250 kgm²

moment of inertia of smaller disc I₂ = 1/2 m r² , m is mass and r is radius of disc . I

I₂ = .5 x 10 x 2.5²

= 31.25 kgm²

3500 rmp = 3500 / 60 rps

n = 58.33 rps

angular velocity of smaller disc ω₂ = 2πn

= 2π x 58.33

= 366.3124 rad /s

applying conservation of angular momentum

I₂ω₂ = ( I₁ +I₂) ω , ω is the common angular velocity

31.25 x 366.3124 = ( 250 +31.25) ω

ω = 40.7 rad / s .

4 0

3 years ago