While the resistance of the variable resistor in the left-hand solenoid is decreased at a constant rate, the induced current thr
ough the resistor RRR will
1 answer:
Answer:
<em>The induced current through resistor R will</em>
<em>b) flow from a to b </em>
Explanation:
The image is shown below, and the full question is written down as
The two solenoids in the figure are coaxial and fairly close to each other. While the resistance of the variable resistor in the left-hand solenoid is decreased at a constant rate, the induced current through the resistor R will
a) Flow from b to a
b) flow from a to b
c) be zero because the rate is constant.
From the image, the current in the left hand solenoid flows from the positive terminal of the battery to the negative terminal in an anticlockwise direction by convention.
Varying a rheostat causes a change in the resistance of electricity through the solenoid, and a changing current through a solenoid will induce current to flow through another solenoid placed nearby. Therefore, the left-hand solenoid induces a current flow on the right-hand solenoid.
Since the current in the left-hand solenoid flows in an anticlockwise direction, then it will have an equivalent magnetic polarity of a north pole on a magnet.
Also remember that Lenz law states that the induce current acts in such a way as to oppose the motion, or action producing it.
In this case, the induced current in the right-hand solenoid will act as to repel the left-hand solenoid away from itself. The only way is by the right-hand solenoid also having a north pole equivalent magnetic pole on it since like poles repel each other. This means that the induced current in the right-hand solenoid will flow in an anticlockwise manner too, from a to b.
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Answer:
Mc = 1920[lb*in]
Explanation:
Para poder solucionar este problema debemos realizar un análisis estático, por tal motivo lo primero es realizar un diagrama de cuerpo libre con las respectivas fuerzas actuando sobre la barra ABC. DE igual manera calcular la geometría de la configuración mostrada.
El diagrama de cuerpo libre se puede ver en la imagen adjunta, con la solución de este problema.
Lo primero es determinar el angulo t, el cual por medio de las propiedades del triangulo rectángulo se puede determinar.
Con este angulo (t) ya determinado, fijamos la atención en el triangulo BCD, este triangulo no es rectángulo, pero por medio de la ley de senos podemos determinar el angulo omega.
Después de determinar el angulo omega, restamos el angulo (t) para poder determinar el angulo (a).
Seguidamente realizamos una sumatoria de momentos alrededor del punto C, utilizado las respectivas fuerzas con los ángulos descompuestos.
El momento en el punto C es de 1920 [Lb*in].
Nota: ya que no se menciona la fuerza en el punto A, esta se desprecia y no se tiene en cuenta en los calculos. En la imagen adjunta se puede ver el procedimiento desarrollado.
