Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 20, Problem 86P

(a)

To determine

The terminal velocity of the rod.

(a)

Expert Solution
Check Mark

Answer to Problem 86P

The terminal velocity of the rod is 3.44m/s_.

Explanation of Solution

Write the expression for motional emf.

    ε=vBL

Here, ε is the motional emf, B is the magnetic field, v is the velocity of the rod and L is the length of the rod.

Write the expression to find the magnitude of the magnetic force.

    FB=IBL

Here, I is the current.

Substitute εR for I in the above equation.

    FB=(εR)BL

Substitute vBL for ε in the above equation.

    FB=(vBLR)BL=vB2L2R

Substitute mg for FB in the above equation and rearrange it to find the velocity v.

    mg=vB2L2Rv=mgRL2B2

Here, m is the mass of the rod and g is the acceleration due to gravity.

Conclusion:

Substitute 0.0150kg for m, 9.80m/s2 for g, 8.00Ω for R, 1.30m for L and 0.450T for B in the above equation.

    v=(0.0150kg)(9.80m/s2)(8.00Ω)(1.30m)2(0.450T)2=3.44m/s

Therefore, the terminal velocity of the rod is 3.44m/s_.

(b)

To determine

The comparison of the rate of gravitational potential energy and the power dissipated in the resistor.

(b)

Expert Solution
Check Mark

Answer to Problem 86P

The rate of gravitational potential energy and the power dissipated in the resistor are the same, 0.505W_.

Explanation of Solution

Write the expression for the rate of change of gravitational energy.

    ΔUΔt=mgΔyΔt

Here, ΔU is the change in potential energy.

Substitute v for ΔyΔt in the above equation.

    ΔUΔt=mgv

Substitute mgRL2B2 for v in the above equation.

    ΔUΔt=mg(mgRL2B2)=m2g2RL2B2                                                                                                   (I)

Write the expression for the power converted in the resistor.

    P=ε2R

Substitute vBL for ε in the above equation.

    P=(vBL)2R

Substitute mgRL2B2 for v in the above equation.

    P=(mgRL2B2BL)2R=m2g2RL2B2                                                                                                     (II)

Conclusion:

Substitute 0.0150kg for m, 9.80m/s2 for g, 8.00Ω for R, 1.30m for L and 0.450T for B in the equation (I).

    ΔUΔt=(0.0150kg)2(9.80m/s2)2(8.00Ω)(1.30m)2(0.450T)2=0.505W

Substitute 0.0150kg for m, 9.80m/s2 for g, 8.00Ω for R, 1.30m for L and 0.450T for B in the equation (II).

    P=(0.0150kg)2(9.80m/s2)2(8.00Ω)(1.30m)2(0.450T)2=0.505W

Therefore, the rate of gravitational potential energy and the power dissipated in the resistor are the same, 0.505W_.

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Chapter 20 Solutions

Physics

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