0
$\begingroup$

While coming across electromagnetic induction, I learned that for an emf to be induced in a coil of wire there should be a change in magnetic flux which is given by $\frac{BA\cos\theta}{t}$ where theta is angle between vector normal to the plane of coil and field lines vector.

But in the case of a single straight wire (not carrying any current) that has a cylindrical shape what should be the vector normal to plane of wire? Can we still apply $BA\cos\theta $ for magetic flux? If yes what should be the theta?

$\endgroup$
1

2 Answers 2

0
$\begingroup$

In your formula theta is the angle between the magnetic field direction at the wire’s location and the direction that is normal (perpendicular) to the plane of one turn in the coil. For example if the coil is a stack of wire circles then the coil’s direction is normal to the plane of any one circle.

Faraday’s law applies to a region of space whether a conducting coil is present or not. Changing magnetic flux through the region induces an electric field that circulates around the boundary of the region. If a conductor happens to be present then that electric field induces current to flow in the conductor.

In the scenario you describe, the conducting cylinder is really just a single circular loop whose normal is directed along the length of the cylinder. The angle you want is the angle between that and the magnetic field. The cylinder must be straight or the angle is not well defined. Also, any induced current will flow in circles, not along the wire’s length.

$\endgroup$
0
0
$\begingroup$

Consider a rectangular loop with dimensions $l$ and $w$. This loop is contained within a magnetic field, which is perpendicular to the plane of the loop.

The loop is then removed from the field with velocity $v$ parallel to the side of length $w$. Therefore the side of length $l$ is moving perpendicular to the field.

In this way, the field, $v$, and side $l$ are all orthogonal.

It can be shown that the emf induced in a conductor of length $l$ moving at velocity $v$ through a field of strength $B$ is $\varepsilon=vBl$, where all three are orthogonal.

Faraday's Law states that the average emf induced due to a change in flux is $\varepsilon=-N\frac{\Delta\Phi}{\Delta t}$. For sake of the argument, we'll assume that $N = 1$ and we'll ignore the negative.

Going back to the loop being removed from the field, we can see that the time it takes for the loop to be removed from the field is $\Delta t = \frac{w}{v}$

The change in flux through the loop as it is removed from the field is $\Delta\Phi=B\Delta A$, where $\Delta A=lw$.

Thus you have from Faraday's Law $\varepsilon=\frac{Blw}{\frac{w}{v}}=vBl$, which gives exactly the equation we stated before.

To introduce the angle $\theta$, we can now consider the same situation, with the plane of the loop offset by angle $\theta$ to its original perpendicular position.

If you now remove the loop with velocity $v$ in the direction of $w$, the component of $v$ perpendicular to the field is $v \cos\theta $. Using the same process as before, you will now see that $\varepsilon=vBl\cos\theta$, and since we have shown this to be equivalent to Faraday's Law, this is there the $\cos\theta$ comes into play in $\varepsilon=\frac{BA\cos\theta}{\Delta t}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.