Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 22, Problem 84P

(a)

To determine

The intensity of the light.

(a)

Expert Solution
Check Mark

Answer to Problem 84P

The intensity of the light is 170W/m2.

Explanation of Solution

Write the equation to find the power of the lamp.

    P=IrmsΔVrms                                                                                           (I)

Here, P is the power of the lamp, Irms is the rms current and ΔVrms is the rms voltage.

Find the equation for the intensity and substitute 2πr2 for A.

    I=PA=P2πr2                                                                                                     (II)

Here, I is the intensity and r is the radius.

Conclusion:

Substitute 12.5A for Irms and 120V for ΔVrms in equation (I) to find P.

    P=(12.5A)(120V)=1500W

Substitute 1500W for P and 1.2m for r in equation (II) to find λ.

    I=1500W2π(1.2m)2=1500W2π(1.44m)=170W/m2

Thus, the intensity of the light is 170W/m2.

(b)

To determine

The energy incident on the face.

(b)

Expert Solution
Check Mark

Answer to Problem 84P

The energy incident on the face is 560J.

Explanation of Solution

Write the equation connecting power, energy and intensity.

    Pface=ΔUfaceΔt=IAface                                                                                       (III)

Here, Pface is the power of the light incident on the face, ΔUface is the energy energy incident on the face, Δt is the time and Aface is the area of the face.

Rearrange and substitute P2πr2 for I to find ΔUface.

    ΔUface=IAfaceΔt=P2πr2AfaceΔt                                                                                      (IV)

Conclusion:

Substitute 1500W for P, 2.8×102m2 for Aface, 2.0min for Δt and 1.2m for r in equation (IV) to find ΔUface.

    ΔUface=1500W2π(1.2m)2(2.8×102m2)(2.0min)=1500W2π(1.2m)2(2.8×102m2)(2.0min)(60s1min)=560J

Thus, the energy incident on the face is 560J.

(c)

To determine

The rms electric and magnetic fields.

(c)

Expert Solution
Check Mark

Answer to Problem 84P

The rms electric field is 250V/m and the rms magnetic field is 8.3×107T.

Explanation of Solution

Write the equation for the rms electric field substituting P2πr2 for I.

    Erms=Iε0c=P2πr21ε0c                                                                                           (V)

Here, Erms is the rms electric field, ε0 is the absolute permittivity and c is the speed of light.

Write the equation for the rms magnetic field.

    Brms=Ermsc                                                                                          (VI)

Here, Brms is the rms magnetic field.

Conclusion:

Substitute 1500W for P, 3.00×108m/s for c, 8.854×1012C2/Nm2 for ε0 and 1.2m for r in equation (V) to find Erms.

    Erms=1500W2π(1.2m)21(8.854×1012C2/Nm2)(3.00×108m/s)=1500W2π(1.44m)1(8.854×1012C2/Nm2)(3.00×108m/s)=250V/m

Substitute 250V/m for Erms and 3.00×108m/s for c in equation (VI) to find Brms.

    Brms=250V/m3.00×108m/s=8.3×107T

Thus, the rms electric field is 250V/m and the rms magnetic field is 8.3×107T.

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Chapter 22 Solutions

Physics

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