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We all know from Ohm's law that $V = IR$; i.e, it can be concluded that $I \propto 1/R$. Also, from resistivity, $R = ρl/A$; i.e., it can be concluded that $R \propto l$.

From the above, we can also conclude that $I \propto 1/l$.

But, after having a look at Lorentz's magnetic force equation, $F = BIl$, I found that $F \propto I$ and $F \propto l$. But our earlier conclusion, $I \propto 1/l$, seems to contradict this.

How? Is there some part of the math or physics here I'm missing?

P.S., I'm studying in tenth grade, so please try not to incorporate extremely complex solutions or explanations in your answers; simplifications to my level of understanding would be greatly appreciated.

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    $\begingroup$ 1. The Lorentz force has not directly to do with Ohms Law. 2. yo use the "proportional wrong. be $\endgroup$ Commented Sep 21, 2025 at 15:06

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You need to be careful about proportionality.

  1. $V=IR \implies I \propto 1/R$ means that if the voltage applied across a resistive circuit element is constant, then the current flowing through it is inversely proportional to its resistance.
  2. $R = \rho \ell/A \implies R \propto \ell$ means that in a uniform rectangular block of material with resistivity $\rho$, the resistance of the block is proportional to its length if its resistivity and cross sectional area are fixed.
  3. Putting those two things together, we can conclude that if we apply a constant voltage across a uniform, rectangular block of resistive material, then the current flowing through the block is inversely proportional to its length.

Now, $F = BI\ell$ gives the magnetic force on a straight wire of length $\ell$ carrying current $I$ in a magnetic field $B$ which is perpendicular to the length of the wire. Therefore,

  1. $F = BI \ell \implies F \propto I$ means that if you hold $B$ and $\ell$ constant, the magnetic force on the wire is proportional to the current flowing through it.
  2. $F = BI\ell \implies F \propto \ell$ means that if you hold $B$ and $I$ constant, the magnetic force on the wire is proportional to its length.

But, our earlier conclusion $I \propto 1/\ell$ seems to contradict this.

There's no contradiction, not least because (1-3) and (4-5) are talking about completely different things. Proportionality isn't typically a universal law relating two quantities - it is generally situation dependent.

For example, the power dissipated in a resistor is given by $P=IV$, where $I$ is the current flowing through the resistor and $V$ is the voltage across it. Using Ohm's law, this becomes $P= I\cdot (IR) = I^2 R$, and the dissipated power is proportional to the resistance.

But I could just as well have substituted $I=V/R$, in which case I would have found that $P = (V/R) \cdot V = V^2/R$, and the dissipated power is inversely proportional to the resistance.

This is an apparent contradiction - how can $P\propto R$ and $P\propto 1/R$? The answer is that $P\propto R$ when the current is held constant. Concretely:

  • If I push 10 mA of current through a collection of different resistors, then the power dissipated in each resistor will be proportional to its resistance. Because $P =I V$ and the current is always the same, the dissipated power is proportional to the voltage. Because $V = IR$, for a given fixed $I$ the voltage increases when we increase $R$.
  • If I apply 10 V across the same collection of resistors, the power dissipated in each resistor will be inversely proportional to its resistance. Because $P = IV$ and the voltage is always the same, the dissipated power is proportional to the current. Because $V = IR$, for a given fixed $V$ the current decreases when we increase $R$.
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    $\begingroup$ This is an excellent answer and explanation. $\endgroup$ Commented Sep 22, 2025 at 17:41
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There is no contradiction. If $I\propto 1/\cal l$ then the total force on the wire is independent of $\cal l$.

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  • $\begingroup$ Yes, if the only impedance in the system comes from the coil, then holding a constant voltage across the coil will result in a force that is independent of coil length. $\endgroup$ Commented Sep 22, 2025 at 14:01
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Your finding of $I\propto L$ from the Lorentz law is based on the premise that, mathematically, if $A\propto B$ and $A\propto C$ it necessarily follows that $B\propto C$. But that is only true if $B$ and $C$ are independent. Here, $B$ and $C$ are current and conductor length, which are not independent.

The root of the problem is you are attempting to compare the current in Ohm's law and the current in Lorentz's law. This is like, as they say, "comparing apples to oranges".

Ohm's law deals with the electric force that drives current through a conductor with resistance due to an electric field that is parallel to the current.

The Lorentz law for a current carrying conductor describes the magnetic force that acts on a conductor as a result of a given amount of current (which is determined by Ohm's law) perpendicular to the magnetic field.

Bottom line: There is no contradiction. Current and conductor length are not independent variables and the two laws involve different forces with different roles.

Hope this helps.

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The Lorentz force has not directly to do with Ohms Law. It is true, that with fixed spezifique resistance , I is proportional to 1/L , but I is not proportional to L in Lorentz force, there I is given or you have to calculate I by Ohms law.So you always have to look for what ist constant in your equation

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In 10th grade, we sure used proportionality a lot to derive out many of the formula we use. Though I personally didn't know about lorentz force at that time, what I did know was newton's force of gravitation formula. The reason why I bring that formula here, is due to the resemblence between this formula and the lorentz one ie. magnetic force on a current carrying wire in a magnetic field (and also the fact that I consider you know the gravitational force formula better, because I did).

We usually do F (attractive force due to gravitation) proportional to product of masses of the objects considering m1 and m2. So, I can consider F proportional to m1 and F proportional to m2. But would that mean, m1 proportional to m2? Obviously not. They are two independent quantities.

Same goes for what you did. When F ∝ I and F ∝ l (in the lorentz formula one), it doesn't mean I ∝ L. Well considering the case when F remains same and B isn't changing then we relate I and L and it does come out that I ∝ 1/L (same as what you figured in V=IR). So, yeah just a little logical error but I would say it's totally because of less exposure to the lorentz formula otherwise, it was a good question.

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