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.

Answer 1

Answer:

external agent exerting the force on the rodexternal agent exerting the force on the rod

Explanation:

external agent exerting the force on the rod

Answer 2

(a) Force is needed to keep the rod sliding at a constant speed v is  $\frac{v}{R}\left[\frac{\mu_{0} i}{2 \pi} \ln \frac{(2 x+l)}{(2 x-l)}\right]^{2}$

(b) Current in the resistance R is $\mathbf{i}^{\prime}=\frac{\mathbf{e}}{\mathbf{R}}=\frac{\mu_{0} \mathbf{v}}{2 \pi R} \ln \frac{(2 x+l)}{(2 x-l)}$

(c) The rate of heat developed in the resistor is

$\frac{1}{R}\left[\frac{\mu_{0} v i}{2 \pi} \ln \frac{(2 x+l)}{(2 x-l)}\right]^{2}$

(d) Power delivered by the external agent exerting the force on the rod is $\frac{1}{R}\left[\frac{\mu_{0} v i}{2 \pi} \ln \frac{(2 x+l)}{(2 x-l)}\right]^{2}$

Explanation:

(a) Here, due to the long wire, the magnetic field B ⃗ assumes that the length of the rod varies.

At a distance a from the wire, we can consider a small element of the length da pin. The magnetic field at distance a is defined by

$\vec{B}=\frac{\mu_{0} i}{2 \pi a}$

Emf caused in rod:

$d e=B v d a$

$d e=\frac{\mu_{0} i}{2 \pi a} \times v \times d a$

Integrating $x-\frac{i}{2} \text { and } x+\frac{l}{2}, \text { we } g e t$

$e=\int_{x-\frac{l}{2}}^{x+\frac{l}{2}} d e$

$=\int_{x-\frac{l}{2}}^{x+\frac{l}{2}} \frac{\mu_{0} i}{2 \pi a} \times v \times d a$

$=\frac{\mu_{0} v i}{2 \pi}\left[\ln \left(x+\frac{l}{2}\right)-\ln \left(x-\frac{l}{2}\right)\right]$

$=\frac{\mu_{0} v i}{2 \pi} \ln \frac{\left(x+\frac{l}{2}\right)}{\left(x-\frac{l}{2}\right)}$

Emf caused by the current-carrying wire into the rod:

$e=\frac{\mu_{0} v i}{2 \pi} \ln \frac{(2 x+l)}{(2 x-l)}$

Now, let the current in the circuit which contains the rod and the resistance be I '.

$i^{\prime}=\frac{e}{R}=\frac{\mu_{0} v i}{2 \pi R} \ln \frac{(2 x+l)}{(2 x-l)}$

Force on the element:

$d F=i^{\prime} B l$

$d F=\frac{\mu_{0} v i}{2 \pi R} \ln \frac{(2 x+l)}{(2 x-l)} \times \frac{\mu_{0} i}{2 \pi a} \times d a$

     $=\left(\frac{\mu_{0} i}{2 \pi}\right)^{2} \frac{v}{R} \frac{(2 x+l)}{(2 x-l)} \frac{d x}{a}$

And

$F=\left(\frac{\mu_{0} i}{2 \pi}\right)^{2} \frac{v}{R} \ln \frac{(2 x+l)}{(2 x-l)} \int_{x-\frac{l}{2}}^{x+\frac{l}{2}} \frac{d a}{a}$

   $=\left(\frac{\mu_{0} i}{2 \pi}\right)^{2} \frac{v}{R} \ln \frac{(2 x+l)}{(2 x-l)} \ln \frac{(2 x+l)}{(2 x-l)}$

   $=\frac{v}{R}\left[\frac{\mu_{0} i}{2 \pi} \ln \frac{(2 x+l)}{(2 x-l)}\right]^{2}$

(b) Current, $\mathrm{i}^{\prime}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{\mu_{0} \mathrm{v}}{2 \pi R} \ln \frac{(2 x+l)}{(2 x-l)}$

(c) The heat rate that is the strength is determined by ,

$w=i^{2} R$

$w=\left[\frac{\mu_{0} v i}{2 \pi R} \ln \frac{(2 x+l)}{(2 x-l)}\right]^{2} R$

  $=\frac{1}{R}\left[\frac{\mu_{0} v i}{2 \pi} \ln \frac{(2 x+l)}{(2 x-l)}\right]^{2}$

(d) The power supplied by the external agency is identical to the heat rate developed.

$p=i^{2} R$

  $=\frac{1}{R}\left[\frac{\mu_{0} v i}{2 \pi} \ln \frac{(2 x+l)}{(2 x-l)}\right]^{2}$