SATHEE: Electro Magnetic Induction And Alternating Currents Question 450

Electro Magnetic Induction And Alternating Currents Question 450

Question: A conducting disc of conductivity a has a radius ‘a’ and thickness ‘f. If the magnetic field B is applied in a direction perpendicular to the plane of the disc changes with time at the rate of $\frac{d B}{d t} = α$ . Calculate the power dissipated in the disc due to the induced current.

Options:

A) $\frac{π t σ a^{4}}{8} α^{2}$

B) $\frac{π t σ a^{4}}{4} α^{2}$

C) $\frac{π t σ a^{4}}{2} α^{2}$

D) $\frac{2 π t σ a^{4}}{3} α^{2}$

Show Answer

Answer:

Correct Answer: A

Solution:

Consider an elemental circle of thickness dr.

The induced emf in the circular path of radius r is

$ε = \frac{d}{d t} (π r^{2} B) = π r^{2} α$

The resistance of circular path is The length of the path being $2 π r$

and tdr is the cross sectional area of current flow.

For the element the power dissipated inside the path is $d P = \frac{ε^{2}}{R} = \frac{π t σ}{2} α^{2} r^{3} d r$

The total dissipated power P is $P = \frac{π t σ}{2} α^{2} \int_{0}^{a} r^{3}$

$d r = \frac{π t σ a^{4}}{8} α^{2}$

Electro Magnetic Induction And Alternating Currents Question 450

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Electro Magnetic Induction And Alternating Currents Question 450

Question: A conducting disc of conductivity a has a radius ‘a’ and thickness ‘f. If the magnetic field B is applied in a direction perpendicular to the plane of the disc changes with time at the rate of dBdt=α . Calculate the power dissipated in the disc due to the induced current.

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu