A metal bar of mass m slides frictionlessly on two parallel conducting rails a distance l apart (Fig. 7.17). A resistor R is connected across the rails, and a uniform magnetic field B, pointing into the page, fills the entire region.

Question

(a) If the bar moves to the right at speed υ, what is the current in the resistor? In
what direction does it flow?

(b) What is the magnetic force on the bar? In what direction?

(c) If the bar starts out with speed [latex]v_{0}[/latex] at time t = 0, and is left to slide, what is its speed at a later time t?

(d) The initial kinetic energy of the bar was, of course [latex]\frac{1}{2} m v_{0}^{2}[/latex] . Check that the energy delivered to the resistor is exactly [latex] \frac{1}{2} m v_{0}^{2}[/latex] .

Accepted Answer

(a) [latex]ε =-\frac{d \Phi}{d t}=-B l \frac{d x}{d t}=-B l v ; ε =I R \Rightarrow I=\frac{B l v}{R}[/latex] (Never mind the minus sign—it just tells you the direction of flow: (v×B) is upward, in the bar, so downward through the resistor.)

(b) [latex]F=I l B=\frac{B^{2} l^{2} v}{R},[/latex] ,to the left.

(c) [latex]F=m a=m \frac{d v}{d t}=-\frac{B^{2} l^{2}}{R} v \Rightarrow \frac{d v}{d t}=-\left(\frac{B^{2} l^{2}}{R m} ight) v \Rightarrow v=v_{0} e^{-\frac{B^{2} l^{2}}{m R} t}[/latex] .

(d) The energy goes into heat in the resistor. The power delivered to resistor is [latex]I^{2} R[/latex] , so

[latex]\frac{d W}{d t}=I^{2} R=\frac{B^{2} l^{2} v^{2}}{R^{2}} R=\frac{B^{2} l^{2}}{R} v_{0}^{2} e^{-2 \alpha t}, ext { where } \alpha \equiv \frac{B^{2} l^{2}}{m R} ; \quad \frac{d W}{d t}=\alpha m v_{0}^{2} e^{-2 \alpha t}[/latex] .

The total energy delivered to the resistor is [latex]W=\alpha m v_{0}^{2} \int_{0}^{\infty} e^{-2 \alpha t} d t=\left.\alpha m v_{0}^{2} \frac{e^{-2 \alpha t}}{-2 \alpha} ight|_{0} ^{\infty}=\alpha m v_{0}^{2} \frac{1}{2 \alpha}=\frac{1}{2} m v_{0}^{2}[/latex] .

Question 7.7: A metal bar of mass m slides frictionlessly on two parallel ...

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