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Sophie [7]
4 years ago
15

What is the relationship between electric currents and magnetic fields?

Physics
1 answer:
Aleksandr [31]4 years ago
7 0
The electric currents induces a magnetic field. They changing magnetic field can be induce an electric current.
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You can jump 3 m (10 feet) off of a diving board and into a swimming pool and be uninjured. However, if you jumped from the same
Sholpan [36]

Answer:

Explanation:

When a person jumps into a swimming pool and is uninjured while the same person is injured when he jumped from the same height onto the concrete floor, this can be explained by the impulse-momentum concept.

When a person jumps into the swimming pool, water provides cushioning and gradually decreases the velocity of a person with a considerable amount of time. While on the other hand concrete floor does not provide the cushioning effect as it is rigid.

So, force is imparted at a short amount of time causing more injury compared to the swimming pool case.

6 0
4 years ago
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.8 cm and a current of 12 A . The bigger l
Stolb23 [73]

Answer:

R_2=6.33cm

Explanation:

The equation for the magnetic field at the center of a circular loop is:

B=\frac{\mu_0I}{2R}

If we name the smaller loop as 1 and the bigger loop as 2 and we impose the same magnitude of the magnetic field at their center, we have:

B_2=B_1

\frac{\mu_0I_2}{2R_2}=\frac{\mu_0I_1}{2R_1}

\frac{I_2}{R_2}=\frac{I_1}{R_1}

R_2=\frac{I_2}{I_1}R_1

Which for our values means:

R_2=\frac{I_2}{I_1}R_1=\frac{(20A)}{(12A)}(3.8cm)=6.33cm

(Notice that we don't need to convert cm to m, we get our result in cm).

3 0
3 years ago
2. A rod 14.0 cm long is uniformly charged and has a total charge of -22.0 μC. Determine the magnitude and direction of the net
Sloan [31]

Explanation:

It is given that,

Length of rod, l = 14 cm = 0.14 m

Total charge, Q=-22\ \mu C=-22\times 10^{-6}\ C

We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m

Electric field at the axis of rod is given by :

E=\dfrac{\lambda}{2\pi \epsilon_o z}

Where

\lambda is the linear charge density, \lambda=\dfrac{Q}{l}

So, E=\dfrac{Q}{2\pi \epsilon_o zl}

E=\dfrac{-22\times 10^{-6}}{2\pi \times 8.85\times 10^{-12}\times 0.36\times 0.14}

E = −7849988.22 N/C

or

E=-7.84\times 10^6\ N/C

Negative sign shows the direction of electric field is inward in all direction. Hence, this is the required solution.

7 0
3 years ago
I Need Help With This
timama [110]

use the equations provided it shouldn't he hard

5 0
3 years ago
Read 2 more answers
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
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