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I came across this question I was unable to solve. Please help. It is a multiple answer type question.

A parallel plat capacitor with plate area A and seperation d is put into flowing conducting liquid of conductivity s. The liquid moves parallel to plates with a constant velocity V. The capacitor is also exposed to external constant magnetic field B directed perpendicular to velocity of the liquid. The capacitor is connected to external circuit R.

a.The current in citcuit is constant

b.magnetic filed causes work which causes heat liberation

c.The power dissipated in external resistance is

d.external forces are to be applied to maintain constant folw of liquid with uniform speed

Jitender Pal , 11 Years ago
Grade 9
anser 1 Answers
Askiitians Tutor Team

Last Activity: 5 Months ago

Let's break down the scenario you've presented regarding the parallel plate capacitor immersed in a conducting liquid and subjected to a magnetic field. This situation involves several physical principles, including electromagnetism, fluid dynamics, and circuit theory. We'll analyze each statement to determine its validity.

Understanding the Setup

In this setup, we have a parallel plate capacitor with plate area A and separation d. The conducting liquid has a conductivity s and flows with a constant velocity V. Additionally, there is an external magnetic field B that is perpendicular to the direction of the liquid's flow. The capacitor is connected to an external resistor R.

Analyzing Each Statement

  • a. The current in the circuit is constant
  • The current in the circuit can be influenced by the movement of the conducting liquid and the induced electromotive force (emf) due to the magnetic field. According to Faraday's law of electromagnetic induction, a changing magnetic field or a conductor moving through a magnetic field can induce an emf. However, if the flow of the liquid is steady and the magnetic field is constant, the induced emf will also be constant, leading to a constant current in the circuit. Thus, this statement is true.

  • b. The magnetic field causes work which causes heat liberation
  • When the conducting liquid flows through the magnetic field, it experiences a Lorentz force, which can do work on the charges in the liquid. This work can lead to energy dissipation in the form of heat due to resistive losses in the liquid and the external resistor. Therefore, this statement is also true.

  • c. The power dissipated in external resistance is
  • The power dissipated in the external resistor can be calculated using the formula P = I²R, where I is the current flowing through the resistor. Since we established that the current is constant, the power dissipation will also be constant over time. The exact value would depend on the current induced by the flow of the liquid and the magnetic field. This statement is true, but it requires further context to specify the exact power value.

  • d. External forces are to be applied to maintain constant flow of liquid with uniform speed
  • To maintain a constant flow of the conducting liquid at uniform speed V, external forces must indeed be applied to counteract any resistive forces, such as viscous drag. If the flow is steady and uniform, it implies that the net force acting on the liquid is zero, which necessitates the application of external forces. Thus, this statement is true as well.

Summary of Findings

Based on our analysis, all four statements regarding the parallel plate capacitor in the conducting liquid under a magnetic field are true. The interplay of the magnetic field, the flow of the liquid, and the electrical circuit creates a fascinating scenario that illustrates key principles of electromagnetism and fluid dynamics.

Should you have any further questions or need clarification on any specific point, feel free to ask!

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