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Question
Consider a situation similar to that of the previous problem except that the ends of the rod slide on a pair of thick metallic rails laid parallel to the wire. At one end the rails are connected by resistor of resistance R. (a) What force is needed to keep the rod sliding at a constant speed v? (b) In this situation what is the current in the resistance R? (c) Find the rate of heat developed in the resistor. (d) Find the power delivered by the external agent exerting the force on the rod.
Solution
(a) Here, the magnetic field \[\overrightarrow B\] due to the long wire varies along the length of the rod.
We will consider a small element of the rod of length da at a distance a from the wire. The magnetic field at a distance a is given by
\[\overrightarrow{B} = \frac{\mu_0 i}{2\pi a}\]
Now,
Induced emf in the rod:-
\[de = Bvda\]
\[de = \frac{\mu_0 i}{2\pi a} \times v \times da\]
Integrating `x-l/2` and `x+l/2,` we get
\[e = \int\limits_{x - \frac{l}{2}}^{x + \frac{l}{2}} de\]
\[ = \int\limits_{x - \frac{l}{2}}^{x + \frac{l}{2}}\frac{\mu_0 i}{2\pi a} vda\]
\[ = \frac{\mu_0 iv}{2\pi}\left[ \ln\left( x + \frac{l}{2} \right) - \ln\left( x - \frac{l}{2} \right) \right]\]
\[ = \frac{\mu_0 iv}{2\pi}\ln\left[ \frac{x + \frac{l}{2}}{x - \frac{l}{2}} \right]\]
Emf induced in the rod due to the current-carrying wire:-
\[e = \frac{\mu_0 iv}{2\pi}\ln\left( \frac{2x + l}{2x - l} \right)\]
Now, let the current produced in the circuit containing the rod and the resistance be i'.
\[i' = \frac{e}{R}\]
\[ = \frac{\mu_0 iv}{2\pi R}\ln\left( \frac{2x + 1}{2x - 1} \right)\]
Force on the element:-
dF = i'Bl
\[\Rightarrow dF = \frac{\mu_0 iv}{2\pi R}\ln\left( \frac{2x + l}{2x - l} \right) \times \left( \frac{\mu_0 i}{2\pi a} \right) \times da\]
\[ = \left( \frac{\mu_0 i}{2\pi} \right)^2 \frac{v}{R}\ln\left( \frac{2x + l}{2x - l} \right)\frac{dx}{a}\]
And,
\[F = \left( \frac{\mu_0 i}{2\pi} \right)^2 \frac{v}{R}\ln\left( \frac{2x + l}{2x - l} \right) \int\limits_{x - \frac{l}{2}}^{x + \frac{l}{2}} \frac{da}{a}\]
\[ = \left( \frac{\mu_0 i}{2\pi} \right)^2 \frac{v}{R}\ln\left( \frac{2x + l}{2x - l} \right)\ln\left( \frac{2x + l}{2x - l} \right)\]
\[ = \frac{v}{R} \left[ \frac{\mu_0 i}{2\pi}\ln\left( \frac{2x + l}{2x - l} \right) \right]^2\]
(b) Current, \[i' = \frac{e}{R} = \frac{\mu_0 iv}{2\pi R}\ln\left( \frac{2x + l}{2x - l} \right)\]
(c) The rate of heat, that is, power, developed is given by
w = i2 R
\[w = \left[ \frac{\mu_0 iv}{2\pi R}\ln\left( \frac{2x + l}{2x - l} \right) \right]^2 R\]
\[ = \frac{1}{R} \left[ \frac{\mu_0 iv}{2\pi}\ln\left( \frac{2x + l}{2x - l} \right) \right]^2\]
(d) Power delivered by the external agency is the same as the rate of heat developed.
Here,
p = i2R
\[= \frac{1}{R} \left[ \frac{\mu_0 iv}{2\pi}\ln\left( \frac{2x + l}{2x - l} \right) \right]^2\]
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