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A particle with charge \(q\) is moving with speed \(v\) in the \(-y\) -direction. It is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=\) \(B_{x} \hat{\imath}+B_{y} \hat{J}+B_{z} \hat{k} .\) (a) What are the components of the force \(\vec{F}\) exerted on the particle by the magnetic field? (b) If \(q>0\), what must the signs of the components of \(\overrightarrow{\boldsymbol{B}}\) be if the components of \(\overrightarrow{\boldsymbol{F}}\) are all nonnegative? (c) If \(q<0\) and \(B_{x}=B_{y}=B_{z}>0\), find the direction of \(\overrightarrow{\boldsymbol{F}}\) and find the magnitude of \(\overrightarrow{\boldsymbol{F}}\) in terms of \(|q|, v\), and \(B_{x}\)

Short Answer

Expert verified
The force is \( q(-vB_{x} \hat{k} + vB_{z} \hat{i}) \). For \( q>0 \), \( B_{z} \geq 0, B_{x} \leq 0 \). For \( q<0, |\vec{F}| = v|q|B_{x}\sqrt{2} \).

Step by step solution

01

Use the Lorentz Force Formula

The force exerted on a charged particle moving in a magnetic field is given by the Lorentz force equation, \( \overrightarrow{\boldsymbol{F}} = q(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \). Since the particle is moving with speed \( v \) in the negative y-direction, we can represent the velocity vector as \( \overrightarrow{\boldsymbol{v}} = -v \hat{j} \).
02

Calculate the Cross Product

Calculate the cross product \( \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}} \). With \( \overrightarrow{\boldsymbol{B}} = B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k} \), the cross product results in:\(-v \hat{j} \times (B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}) = -vB_{x} \hat{k} + vB_{z} \hat{i} \). Hence, \( \overrightarrow{\boldsymbol{F}} = q(-vB_{x} \hat{k} + vB_{z} \hat{i}) \).
03

Determine Force Components for q>0

a) The components of force \( \overrightarrow{\boldsymbol{F}} \) are:\( F_{x} = qvB_{z}, \, F_{y} = 0, \, F_{z} = -qvB_{x} \).b) For all force components to be nonnegative when \( q > 0 \), we need:\( B_{z} \geq 0 \) and \( -B_{x} \geq 0 \) which implies \( B_{x} \leq 0 \).
04

Analyze the Force for q

c) If \( q<0 \) and \( B_{x}=B_{y}=B_{z}>0 \), substitute these values into the force components:\( F_{x} = qvB_{z}, \, F_{y} = 0, \, F_{z} = -qvB_{x} \). With \( q<0 \), both \( qvB_{z} \) and \( -qvB_{x} \) are positive (since \( |q| \) means \( q = -|q| \)). Thus, the force \( \overrightarrow{\boldsymbol{F}} \) points in the positive x- and z-directions.
05

Calculate the Magnitude of the Force

To find the magnitude of \( \overrightarrow{\boldsymbol{F}} \) for \( q<0 \), use the magnitude formula:\[ |\overrightarrow{\boldsymbol{F}}| = \sqrt{F_{x}^2 + F_{y}^2 + F_{z}^2} = \sqrt{(v \cdot |q| B_{x})^2 + (v \cdot |q| B_{x})^2} \].Simplify:\[ |\overrightarrow{\boldsymbol{F}}| = v \cdot |q| \cdot B_{x} \sqrt{2} \].

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Key Concepts

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

Magnetic Force
The concept of magnetic force is integral in understanding how charged particles behave in a magnetic field. When a charged particle, such as an electron or proton, enters a magnetic field, it experiences a force known as the magnetic force. This force does not act in the direction of the magnetic field or the particle's velocity, but rather it acts perpendicular to both.
According to the Lorentz Force Law, the magnitude of the magnetic force (\(F\)) is determined by the equation:\[F = q(v \times B)\], where \(q\) is the charge of the particle, \(v\) is the velocity, and \(B\) is the magnetic field. Remember, this equation is a vector equation, denoting both the direction and magnitude of the force.
The direction of the magnetic force is given by the right-hand rule, which is a method for determining the direction of a cross product. Keep in mind that this force is always perpendicular to the velocity of the particle, meaning it does no work on the particle. As a result, a magnetic field can change the direction of a particle's velocity, but it cannot change its speed.
Cross Product
The vector cross product is a crucial mathematical operation used to compute the magnetic force in physics. It involves two vectors and results in a third vector that is perpendicular to the plane formed by the original vectors. In the context of magnetic force, the vectors involved are the velocity vector \(\overrightarrow{v}\) of a charged particle and the magnetic field vector \(\overrightarrow{B}\).
When carrying out the cross product \(\overrightarrow{v} \times \overrightarrow{B}\), use the determinant method or the right-hand rule. You align your hand according to the direction of the first vector (velocity, in this case), and curl your fingers towards the second vector (magnetic field). Your thumb then points in the direction of the resulting vector (magnetic force).
If the vectors are represented in the cartesian coordinates with components, the cross product can be expanded as:\(\overrightarrow{v} \times \overrightarrow{B} = (v_x \hat{i} + v_y \hat{j} + v_z \hat{k}) \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})\). This mathematical tool helps define both the magnitude and direction of the magnetic force, crucial for determining how charged particles move within a magnetic field.
Charged Particle in Magnetic Field
When a charged particle moves through a magnetic field, it undergoes a force that affects its trajectory. This interaction is pivotal for many applications, such as in particle accelerators or magnetic confinement in fusion reactors.
The path of a charged particle in a uniform magnetic field is typically circular or helical. This is due to the perpendicular nature of the magnetic force in relation to the particle's velocity. For a positive charge, the magnetic force direction can be predicted using the right-hand rule. Conversely, for a negative charge, the direction is reversed.
Since the force is always perpendicular to the velocity, it acts as a centripetal force, leading the particle along a curved path without changing its speed. The radius of this path is dependent on the charge, velocity, mass of the particle, and the strength of the magnetic field, determined by:\[r = \frac{mv}{|q|B}\]. Here, \(m\) is mass, \(v\) is velocity, \(q\) is charge, and \(B\) is the magnetic field strength.
  • Velocity vectors align due to this force.
  • Charged particles can be deflected, following a predictable path.
  • The energy of the particle remains constant as the magnetic field does no work.
Physics for JEE
In the context of Physics exams such as JEE, understanding the principles of charged particles in magnetic fields is crucial. The examination tests how well students grasp these core concepts of electromagnetism, which are foundational to physics.
Key focuses of JEE include:
  • Applying the Lorentz force formula effectively to solve problems.
  • Evaluating vector products to determine force directions.
  • Using mathematical skills to calculate magnitudes of forces.
For JEE success, students need a strong understanding of how charged particles interact with magnetic fields, how to apply vector mathematics like the cross product, and how to interpret the resulting forces and motion.
Questions in exams often involve step-by-step derivations, requiring students to demonstrate not just knowledge, but also their ability to apply the Lorentz force law. The focus is on both conceptual understanding and practical problem solving, preparing students for more advanced studies in physics.

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Most popular questions from this chapter

A beam of protons traveling at \(1.20 \mathrm{~km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. E7.15). The beam travels a distance of \(1.18 \mathrm{~cm}\) while in the field. What is the magnitude of the magnetic field?

A thin, uniform rod Figure \(P 7.45\) with negligible mass and length \(0.200\) \(\mathrm{m}\) is attached to the floor by less hinge at point \(P\) (Fig. horizontal spring with force \(k=4.80 \mathrm{~N} / \mathrm{m}\) connects the of the rod to a vertical wall. in a uniform magnetic field \(0.340 \mathrm{~T}\) directed into the plane figure. There is current \(I=6.50 \mathrm{~A}\) in the rod, in the direction shown. (a) Calculate the torque due to the magnetic force on the rod, for an axis at \(P\). Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? Explain. (b) When the rod is in equilibrium and makes an angle of \(53.0^{\circ}\) with the floor, is the spring stretched or compressed? (c) How much energy is stored in the spring when the rod is in equilibrium?

(a) An \({ }^{16} \mathrm{O}\) nucleus (charge \(+8 e\) ) moving horizontally from west to east with a speed of \(500 \mathrm{~km} / \mathrm{s}\) experiences a magnetic force of \(0.00320 \mathrm{nN}\) vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, \(2.10-\mathrm{T}\) magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be \(4.60 \mathrm{pN}\), vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction.

A \(150-g\) ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125-\mathrm{m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude \(0.250 \mathrm{~T}\) and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section \(8.5\) ), in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times\) \(10^{6} \mathrm{~m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\). (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

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