EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 28, Problem 83P
To determine

The angle with the equilibrium position will vary with time according to θ0et/2πcosωt .

Expert Solution & Answer
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Answer to Problem 83P

The angle with the equilibrium will vary according to relation θ0et/2πcosωt .

Explanation of Solution

Given data:

The uniform horizontal magnetic field is B .

The angle θ is 0 .

The value of τ is RI/(NBA)2 .

The value of ω is κI .

The value of ω is ω012( ω 0 τ)2 .

Formula:

The expression for the restoring torque on the coil is given by,

  τrestoring=κθ

The expression for the retarding torque is given by,

  τrestarding=NiBAcosθ

The expression for the net torque on the coil is expressed as,

  τ=Iα

The expression for the induced emf in the coil is evaluated as,

  ε=Nddt(BAsinθ)=NBAcosθdθdt

The expression for the induced current in the coil is calculated as,

  i=|ε|R

Calculation:

The net torque acting on the coil is calculated as,

  τ=Iατrestoringτrestarding=Id2θdt2κθNiBAcosθ=Id2θdt2

The magnitude of the current through the coil is calculated as,

  i=|ε|R=|NBAcosθ dθ dt|R=NBAcosθ dθ dtR

The expression for the net torque acting on the torque is further evaluated as,

  i=|ε|R=|NBAcosθ dθ dt|R=NBAcosθ dθ dtR

The expression for the differential is given by,

  κθNiBAcosθ=Id2θdt2κθN( NBAcosθ dθ dt R)BAcosθ=Id2θdt2κθI ( NBA )2RId2θdt2=Id2θdt2

The above equation is evaluated by κI for ω02 and ( NBA)2RI for 1τ in the above equation,

  d2θdt2+1τdθdt+ω02=0 …… (I)

The solution for the above differential equation is given by,

  θ(t)=θ0et/2πcosωt

The differential equation dθdt is evaluated as,

  dθdt=d[ θ 0 e t/ 2π cos ω t]dtdθdt=θ0et/2π[ωsinωt+cos ω t2τ]

The expression for d2θdt2 is evaluated as,

  d2θdt2=ddt(θ0e t/ 2π [ ω sin ω t+ cos ω t 2τ])=ddt(θ0e t/ 2π [ ω sin ω t+ cos ω t 2τ])=ddt(θ0e t/ 2π [ ω sin ω t+ cos ω t 2τ])=θ0et/2π(ωcosωt ω sin ω t 2τ)=θ0et/2π( ω 2cosωt ω τsinωt+ ω cos ω t 4 τ 2 )

Solve further as,

  d2θdt2=θ0et/2π(ω2cosωtωτsinωt+ωcosωt4τ2)

The values of d2θdt2 and dθdt in equation (I) are evaluated as,

  [ θ 0 e t/ 2π ( ω 2 cos ω t ω τ sin ω t+ ω cos ω t 4 τ 2 )+ 1 τ( θ 0 e t/ 2π [ ω sin ω t+ cos ω t 2τ ])+ ω 0 2]=0θ0et/2π( ω 2cosωt+ ω 4τcosωtω0cosωt)=0cosωt( ω 2+1 4τω0)=0ω=ω0( 1 1 ( 2τ ω 0 ) 2 )

Conclusion:

Therefore, the angle with the equilibrium will vary according to relation θ0et/2πcosωt .

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Chapter 28 Solutions

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