Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 31, Problem 67AP

(a)

To determine

The induced emf in the closed loop POQ.

(a)

Expert Solution
Check Mark

Answer to Problem 67AP

The emf induced in the loop is 0.125V in clockwise direction.

Explanation of Solution

Write the expression for emf induced in the coil due to change in magnetic field.

    ε=dΦBdt                                                                                                                 (I)

Here, ε is the induced emf, ΦB is the magnetic flux through loop and t is the time taken.

Write the expression for magnetic flux through loop.

    ΦB=BANcosθ                                                                                                      (II)

Here, B is the magnetic field, N is the number of turns in the coil, θ is the angle between the magnetic field and the normal vectors of the loop and A is the area of the loop.

Substitute 1 for N, θa22 for A and 0° for θ in equation (II) to calculate ΦB.

    ΦB=B(θa22)(1)cos0°=B(θa22)

Substitute B(θa22) for ΦB in equation (I) to calculate ε.

    ε=d(B(θa22))dt=Bd(θa22)dt=Ba22(dθdt)                                                                                                 (III)

Write the expression for angular velocity.

    ω=dθdt                                                                                                                  (IV)

Here, ω is the angular velocity.

Substitute ω for dθdt in equation (III) to calculate ε.

    ε=Ba22(ω)                                                                                                           (V)

Conclusion:

Substitute 50.0cm for a, 0.500T for B and 2.00rad/s for ω in equation (V) to calculate ε.

    ε=(0.500T)(50.0cm(1m1×102cm))22(2.00rad/s)=(0.500T)(0.5m)22(2.00rad/s)=0.252V=0.125V

Therefore, the emf induced in the loop is 0.125V in clockwise direction.

(b)

To determine

The induced current in the loop POQ.

(b)

Expert Solution
Check Mark

Answer to Problem 67AP

The induced current in the loop is 0.02A.

Explanation of Solution

The following figure shows the loop POQ and the length of the sides.

Physics for Scientists and Engineers With Modern Physics, Chapter 31, Problem 67AP

Figure-(1)

From figure (1) it is shown that the length of the arc PQ is lθ and length of the sides PO and OQ is l.

Rewrite equation (IV) for θ.

    ω=θtθ=ωt                                                                                                                    (VI)

Total length of the loop is,

    L=OP+OQ+PQ

Substitute lθ for PQ and l for PO and OQ in the above equation.

    L=l+l+lθ                                                                                                           (VII)

Resistance per length for the conductor is given as.

    RL=5.00Ω/m                                                                                                      (VIII)

Write the expression for induced current in the loop.

    I=εR                                                                                                                      (IX)

Here, I is the induced current.

Conclusion:

Substitute 2.00rad/s for ω and 0.250s for t in Equation (VI) to calculate θ.

    θ=(2.00rad/s)(0.250s)=0.5rad

Substitute 50.0cm for l and 0.5rad for θ in Equation (VII) to calculate L.

    L=[(50.0cm)+(50.0cm)+(50.0cm)(0.5rad)](1m1×102cm)=(0.5m)+(0.5m)+(0.5m)(0.5rad)=1m+0.25m=1.25m

Substitute 1.25m for L in Equation (VIII) to calculate R.

    R1.25m=5.00Ω/mR=(5.00Ω/m)(1.25m)R=6.25Ω

Substitute 0.125V for ε and 6.25Ω for R in Equation (IX) to calculate I.

    I=0.125V6.25Ω=0.02A

Therefore, the induced current in the loop is 0.02A.

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Chapter 31 Solutions

Physics for Scientists and Engineers With Modern Physics

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What is Electromagnetic Induction? | Faraday's Laws and Lenz Law | iKen | iKen Edu | iKen App; Author: Iken Edu;https://www.youtube.com/watch?v=3HyORmBip-w;License: Standard YouTube License, CC-BY