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murzikaleks [220]
3 years ago
7

A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

icular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.220 T/s.
Required:
What is the magnitude of the electric field induced in the ring?
Physics
1 answer:
labwork [276]3 years ago
8 0

Answer:

Ein: 2.75*10^-3 N/C

Explanation:

The induced electric field can be calculated by using the following path integral:

\int E_{in} dl=-\frac{\Phi_B}{dt}

Where:

dl: diferencial of circumference of the ring

circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m

ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)

The electric field is always parallel to the dl vector. Then you have:

E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)

Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:

E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}

hence, the induced electric field is 2.75*10^-3 N/C

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Alika [10]

Answer:

yes

Explanation:

3 0
2 years ago
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Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo
Naily [24]

Answer:

t= 1,56 s ,   x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

        y = y₀ + v_{oy} t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

       0 = y₀ – ½ g t²

       t= √ (2 y₀/g)

calculemos

       t= √ ( 2 12 / 9,8)

       t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida  

        x = v₀ₓ t

        x = 80 1,56

        x= 124,8 m

La velocidad de impacto cuando toca el suelo

        vx = v₀ₓ = 80 ms

        v_{y} = v_{oy} – gt

        v_{y} = - 9,8 1,56

        v_{y} = - 15,288 m/s

la velocidad es

       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

        y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

       0 =y₀I - ½ g t²

       t = √ (2y₀ / g)

let's calculate

       t = √ (2 12 / 9.8)

       t = 1.56 s

Projectile range is the horizontal distance traveled

        x = v₀ₓ t

        x = 80 1.56

        x = 124.8 m

Impact speed when it hits the ground

        vₓ = v₀ₓ = 80 ms

        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

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A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr
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