Solve the differential equation of motional emf

[christang_1023]

Homework Statement

A bar of mass m, length d, and resistance R slides without friction in a horizontal plane, moving on parallel rails as shown below. A battery that maintains a constant emf ε is connected between the rails, and a constant external magnetic field B is directed perpendicular to the plane of the page. Assuming the bar starts from rest and that the field produced by the rails is very small compared to the constant external magnetic field, show that at time t it moves with what speed

Relevant Equations

Shown below

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Above is the figure of the problem.

I am trying to solve x(t) and differentiate it to obtain v(t); however, I have difficulty solving the differential equation shown below.

$$ v(t)=\int a(t)dt=\int \frac{B(\varepsilon-Blv)d}{Rm}dt \Rightarrow \frac{dx}{dt}=\frac{B\varepsilon d}{Rm}t-\frac{Bld}{Rm}x $$

 

[haruspex]

Assuming your ODE is correct, try substituting x=y+αt+β and see if there are values for α and β which simplify the equation.

 

[PeroK]

Problem Statement
A bar of mass m, length d, and resistance R slides without friction in a horizontal plane, moving on parallel rails as shown below. A battery that maintains a constant emf ε is connected between the rails, and a constant external magnetic field B is directed perpendicular to the plane of the page. Assuming the bar starts from rest and that the field produced by the rails is very small compared to the constant external magnetic field, show that at time t it moves with what speed
Relevant Equations
Shown below

View attachment 242527.
Above is the figure of the problem.

I am trying to solve x(t) and differentiate it to obtain v(t); however, I have difficulty solving the differential equation shown below.

$$ v(t)=\int a(t)dt=\int \frac{B(\varepsilon-Blv)d}{Rm}dt \Rightarrow \frac{dx}{dt}=\frac{B\varepsilon d}{Rm}t-\frac{Bld}{Rm}x $$

If you are trying to find the speed, why not solve for that directly?

What is $l$? Is that the same as $d$?

 

[christang_1023]

If you are trying to find the speed, why not solve for that directly?

What is $l$? Is that the same as $d$?

$l$ is the same as $d$.
Not solving for v directly was because I consider that $v(t)=v_0-\int a(t)dt,$ which contains $\int vdt=x$; therefore, I wanted to solve the ODE for x first, which is shown above.

Fortunately, thanks to your hint, I manage to solve v directly based on Newton's 2nd law:
$$ F=m \frac{dv}{dt} \Rightarrow \frac{B\varepsilon d-B^2d^2v}{mR}=\frac{dv}{dt} $$

 

[PeroK]

$l$ is the same as $d$.
Not solving for v directly was because I consider that $v(t)=v_0-\int a(t)dt,$ which contains $\int vdt=x$; therefore, I wanted to solve the ODE for x first, which is shown above.

Fortunately, thanks to your hint, I manage to solve v directly based on Newton's 2nd law:
$$ F=m \frac{dv}{dt} \Rightarrow \frac{B\varepsilon d-B^2d^2v}{mR}=\frac{dv}{dt} $$

What does your answer look like if $B$ is small? Does it make sense?

 

[christang_1023]

What does your answer look like if $B$ is small? Does it make sense?

The result is $$ v(t)=\frac{\varepsilon}{Bd} (1-e^{-\frac{B^2d}{mR}t}) $$
According to L'Hospital's Rule, $\lim_{B \to 0} v(t)=0,$ that is to say when B is sufficiently small, the bar won't move at all, and the whole circuit remains still. It is reasonable.