# Magnetic field and inductor

## Homework Statement ## The Attempt at a Solution

In this problem as the rod moves towards right, magnetic force of magnitude ilB starts acting towards left on the rod.This decreases the speed of the rod .The EMF induced in the rod is Blv where v is the instantaneous speed of the rod . As speed of the rod decreases , the induced EMF also decreases .

If rod were moving at constant speed , the current through the inductor would increase with time .But since EMF decreases with time , the current should decrease .

How should I take the two factors into account ?

Thanks

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TSny
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Can you write an equation that relates the current (or rate of change of current) in the inductor to the EMF?

Can you write an equation that relates the current (or rate of change of current) in the inductor to the EMF?

E = Ldi/dt

Blv = Ldi/dt , v is the instantaneous speed of the rod .

cnh1995
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How should I take the two factors into account ?
No resistance, no dissipation of energy. What can you say about the motion then?

TSny
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E = Ldi/dt

Blv = Ldi/dt , v is the instantaneous speed of the rod .
Good. What equation governs the motion of the rod?

(Note that two of the multiple choice answers can be eliminated just based on dimensional analysis.)

What equation governs the motion of the rod?

dv/dt = -ilB/m

TSny
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dv/dt = -ilB/m
Right.

Do I need to analyze the equation ##\ddot{i} + \frac{B^2l^2}{m}i=0## so as to get the maximum value of i ?

rude man
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IMO it's a bit dangerous to try to reason this system out verbally.
Get two ODE's, one in current and one in velocity, eliminate velocity & wind up with one ODE in current.

TSny
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Do I need to analyze the equation ##\ddot{i} + \frac{B^2l^2}{m}i=0## so as to get the maximum value of i ?
That should work. [Edit: But see rude man's post below regarding a missing factor of L somewhere.]

[ @cnh1995 has hinted at another approach that can get the answer very quickly. You might go ahead and work it out using your differential equation for i. Then try to see this other approach.]

rude man
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Do I need to analyze the equation ##\ddot{i} + \frac{B^2l^2}{m}i=0## so as to get the maximum value of i ?
Wow, you're quick! But put in the missing L.

• TSny
[ @cnh1995 has hinted at another approach that can get the answer very quickly. You might go ahead and work it out using your differential equation for i. Then try to see this other approach.]

This is how I worked out the problem in first attempt . But then I tried to analyse the problem further by thinking about how current was varying. I didn't mention it in the OP so as to not distract others and focus primarily on how things were happening in this problem .

I will try and see if I can get the same result from the DE .

TSny
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This is how I worked out the problem in first attempt . But then I tried to analyse the problem further by thinking about how current was varying. I didn't mention it in the OP so as to not distract others and focus primarily on how things were happening in this problem .

I will try and see if I can get the same result from the DE .
OK. Great.

cnh1995
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This is how I worked out the problem in first attempt Yes, using a simple energy conservation equation, you can mentally work it out.

But I agree the DE approach is better for understanding as you'll get to solve a second order DE and you could see how current and velocity change w.r.t. each other.

• Vibhor
rude man
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Cons. of energy works to answer the problem but does it reveal the oscillatory motion? Which is why I like solving the ODE.

Interesting that B plays no part & could =0. EDIT: except that the oscillation frequency is proportional to B. B here is the external mag field, not that generated by the loop.

T's "spring constant" k = B2l2/(L) so ω = √(k/m) as in the mechanical analog.

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TSny
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Cons. of energy works to answer the problem but does it reveal the oscillatory motion? Which is why I like solving the ODE.

Interesting that B plays no part & could =0.
Yes, it is interesting.

To get the answer from the energy approach, you need to argue that there will be a time (or times) for which the rod has zero velocity so that all of the initial KE is converted into magnetic energy in the inductor. If B = 0, then of course there will not be any current at all.

It's like a mass attached to a horizontal spring and given an initial velocity when the spring is unstretched. As long as the spring constant is greater than zero, then all of the initial KE will eventually be converted at some time (or times) into elastic potential energy and the value of the spring constant is irrelevant as regards the maximum PE. But if the spring constant is zero, there will be no conversion of KE to PE.

• Vibhor
Assuming i = imcos(ωt+Φ) , and using the initial conditions(at t=0) i =0 and di/dt = Blv0/L , I get Φ = 3π/2 .

Φ = π/2 gives an extra minus sign .

Thus , i = imsinωt

Is that correct ?

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TSny
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Assuming i = imcos(ωt+Φ) , and using the initial conditions(at t=0) i =0 and di/dt = Blv0/L , I get Φ = 3π/2 .

Φ = π/2 gives an extra minus sign .

Thus , i = imsinωt

Is that correct ?
Looks good. The sign of i just depends on your choice of positive direction for the current.

Solving a DE is far more interesting and insightful than COE .Isn't it ?

Is this the only way to find the solution of the DE in this particular problem or we could also have guessed the solution . Could you have straightaway said that solution would be of the form imsinωt ?

Now if I want to analyze the motion of the rod then equation would be dv/dt = -(Biml)sinωt/m .

Solving it further gives ,

Speed of the rod v = v0 + (Biml)/(mω)cosωt .

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TSny
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Now if I want to analyze the motion of the rod then equation would be dv/dt = -(Biml)sinωt/m .

Solving it further gives ,

Speed of the rod v = v0 + (Biml)/(mω)cosωt .
This isn't correct. Note that at t = 0, you don't get v = v0. It might be easier to get v by using ε = Bvl and ε = Ldi/dt.

• Vibhor
TSny
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Solving a DE is far more interesting and insightful than COE .Isn't it ?
I agree.

Is this the only way to find the solution of the DE in this particular problem or we could also have guessed the solution . Could you have straightaway said that solution would be of the form imsinωt ?
I guess there are ways to see that the motion of the rod must be SHM without setting up the explicit DE.

From di/dt = Blv/L you get

i = (Bl/L) ∫vdt = (Bl/L) Δx, where Δx is the displacement from the starting position.

The force on the rod is then F = -ilB = -k Δx for a constant k. Thus, SHM.

• Vibhor and cnh1995
It might be easier to get v by using ε = Bvl and ε = Ldi/dt.

Placing the origin at the initial position of the rod I get displacement x of the rod = (Limsinωt)(Bl) = (Li/Bl) . Is that correct ?

Edit : I see this is the same result you are getting in the above post in a much more elegant way .I first differentiated "i" to get the expression for "v" and then again used integration to get expression for the displacement .

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This isn't correct. Note that at t = 0, you don't get v = v0.

From dv/dt = -(Biml)sinωt/m .

Solving it gives ,

Speed of the rod v = v0 + (Biml)(cosωt-1)/(mω)

Is it correct now ?

This expression looks different from the earlier expression I got which gave correct value of displacement . It seems I am messing up somewhere .

TSny
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From dv/dt = -(Biml)sinωt/m .

Solving it gives ,

Speed of the rod v = v0 + (Biml)(cosωt-1)/(mω)

Is it correct now ?
It's correct, but it will simplify nicely if you substitute for ω and im in terms of the given quantities: v0, B, L, l, and m.

It's much easier to get v from your solution for i by using vBl = Ldi/dt.