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Equal magnitudes Suppose we have a situation in which the component of the magnetic field parallel to the plane of a sheet has the same magnitude on both sides, but changes direction by \(90^{\circ}\) in going through the sheet. What is going on here? Would there be a force on the sheet? Should our formula for the force on a current sheet apply to cases like this?

Short Answer

Expert verified
There would be no force on the sheet due to the change in direction of the magnetic field by \(90^{\circ}\). Our force formula for a current sheet may not apply due to this direction change.

Step by step solution

01

Analyze the magnetic field

Understanding magnetic fields is crucial. In this scenario, the magnetic field's component stays the same in magnitude but changes direction by \(90^{\circ}\). Magnetic fields do change their directions when they are influenced by materials that contain electric charges such as the current sheet mentioned.
02

Force Evaluation

There wouldn't be any force on the sheet due the magnetic field in this scenario. The reason is that the force on a current-carrying conductor (or this current sheet in the scenario) due to a magnetic field depends not only on the field strength (magnitude) but also the angle between the direction of the field and the current. If the magnetic field is perpendicular to the current (which can be inferred by the \(90^{\circ}\) change), the magnetic force is zero.
03

Formula applicability

Our formula for the force on a current sheet may not apply to this case, since it assumes a constant magnetic field direction, while here we have a significant change across the sheet boundary. Moreover, this formula is derived under the assumption of a linear relationship between the field and the force, which may not hold true for a changing direction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Force
In physics, the concept of magnetic force is essential when studying how magnets and electromagnetic fields interact with materials. The magnetic force that acts on a current-carrying conductor is determined by the magnetic field in which it is placed. If you have two pieces of information: the direction of the magnetic field and the direction of the current, you can predict if a force will act on the sheet.
The force is calculated using the formula: \( F = I \times L \times B \times \sin(\theta) \), where:
  • \(F\) is the magnetic force.
  • \(I\) is the current flowing through the conductor.
  • \(L\) is the length of the conductor in the magnetic field.
  • \(B\) is the magnetic field strength.
  • \(\theta\) is the angle between the direction of the magnetic field and the current.
Change in any of these components, particularly \(\theta\), significantly impacts the force experienced.
Current Sheet
A current sheet is a thin, flat layer in which electric currents flow. It plays a pivotal role in understanding how magnetic fields influence materials containing electric charges. In this scenario, the current sheet is placed where the magnetic field remains constant in magnitude but changes direction by 90 degrees.
When electric charges move through the current sheet, the resulting magnetic field changes can affect everything from the turning of a compass needle to the functioning of electrical devices. In these sheets, electric charges usually travel in paths that generate their own magnetic fields, which interact with external influences.
Direction Change
A change in the direction of a magnetic field is a significant factor in determining its interactions with current sheets. If a magnetic field changes direction sharply, such as a 90-degree turn through a material, this can lead to an analysis of forces acting on said material.
Direction change in magnetic fields highlights the importance of vector quantities where both direction and magnitude are essential. In our scenario, the significant change in the magnetic field's direction implies that it becomes perpendicular to the current sheet. When the field is perpendicular, it suggests minimal or no magnetic force acting upon the sheet. This understanding is crucial for designing and analyzing systems where orientation adjustments are made, such as in motors or generators.
Force Evaluation
Evaluating the force acting on a current sheet involves understanding the relationship between the magnetic field direction, current direction, and magnetic force. In cases where the magnetic field is perpendicular to the current, as implied by a 90-degree direction change, the magnetic force is zero.
This evaluation considers the principle that force is only exerted when there is an effective component of the magnetic field working parallel or at an angle less than 90 degrees to the current. Thus, understanding this principle can help in scenarios where reducing unwanted forces is crucial, like in magnetic levitation or precise electronic instruments.
Evaluating forces accurately also informs us about situations where our typical formula for magnetic force on a current sheet may not apply due to changes that disrupt linear relations between involved quantities.

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Most popular questions from this chapter

Hall voltage \(* *\) A Hall probe for measuring magnetic fields is made from arsenic-doped silicon, which has \(2 \cdot 10^{21}\) conduction electrons per \(\mathrm{m}^{3}\) and a resistivity of \(0.016 \mathrm{ohm}-\mathrm{m}\). The Hall voltage is measured across a ribbon of this \(n\)-type silicon that is \(0.2 \mathrm{~cm}\) wide, \(0.005\) \(\mathrm{cm}\) thick, and \(0.5 \mathrm{~cm}\) long between thicker ends at which it is connected into a \(1 \mathrm{~V}\) battery circuit. What voltage will be measured across the \(0.2 \mathrm{~cm}\) dimension of the ribbon when the probe is inserted into a field of 1 kilogauss?

Integral of \(A\), flux of \(B\) Show that the line integral of the vector potential \(\mathbf{A}\) around a closed curve \(C\) equals the magnetic flux \(\Phi\) through a surface \(S\) bounded by the curve. This result is very similar to Ampère's law, which says that the line integral of the magnetic field \(\mathbf{B}\) around a closed curve \(C\) equals (up to a factor of \(\mu_{0}\) ) the current flux \(I\) through a surface \(S\) bounded by the curve.

Field at the tip of a cone \(* *\) A hollow cone (like a party hat) has vertex angle \(2 \theta\), slant height \(L\), and surface charge density \(\sigma\). It spins around its symmetry axis with angular frequency \(\omega\). What is the magnetic field at the tip?

Field in the plane of a ring ** A ring with radius \(R\) carries a current \(I\). Show that the magnetic field due to the ring, at a point in the plane of the ring, a distance \(a\) from the center (either inside or outside the ring), is given by $$ B=2 \cdot \frac{\mu_{0} I}{4 \pi} \int_{0}^{\pi} \frac{(R-a \cos \theta) R d \theta}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}} $$ Hint: The easiest way to handle the cross product in the BiotSavart law is to write the Cartesian coordinates of \(d l\) and \(\mathbf{r}\) in terms of an angle \(\theta\) in the ring. This integral can't be evaluated in closed form (except in terms of elliptic functions), but it can always be evaluated numerically if desired. For the special case of \(a=0\) at the center of the ring, the integral is easy to do; verify that it yields the result given in

Fields from two rings \(*\) A ring with radius \(r\) and linear charge density \(\lambda\) spins with frequency \(\omega\). A second ring with radius \(2 r\) has the same density \(\lambda\) and frequency \(\omega\). Each ring produces a magnetic field at its center. How do the magnitudes of these fields compare?

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