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A \(1.00-\mathrm{kg}\) ball having net charge \(Q=5.00 \mu \mathrm{C}\) is thrown out of a window horizontally at a speed \(v=20.0 \mathrm{~m} / \mathrm{s}\). The window is at a height \(h=20.0 \mathrm{~m}\) above the ground. A uniform horizontal magnetic field of magnitude \(B=\) \(0.0100 \mathrm{~T}\) is perpendicular to the plane of the ball's trajectory. Find the magnitude of the magnetic force acting on the ball just before it hits the ground. (Hint: Ignore magnetic forces in finding the ball's final velocity.)

Short Answer

Expert verified
The magnitude of the magnetic force acting on the ball just before it hits the ground is \(F = QvB = 1.00 N\).

Step by step solution

01

Calculate the time the ball is in the air

First need to calculate the time the ball is in the air before it hits the ground. Use the formula \(t=\sqrt{\frac{2h}{g}}\) where \(h = 20 m\) is the height and \(g=9.81 m/s^2\) is the acceleration due to gravity.
02

Find the final horizontal speed

Since no horizontal forces are acting on the ball, its horizontal speed remains the same, i.e., \(v = 20 m/s\).
03

Calculate the magnetic force

To get the magnetic force acting on the ball, use the formula \(F = QvB\), where \(Q = 5.00 \mu C\) is the charge on the ball, \(v = 20 m/s\) is the velocity, and \(B = 0.01 T\) is the magnetic field. Here, the angle between the velocity and the magnetic field is 90 degrees, so the force is maximized.

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Key Concepts

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

Motion in Magnetic Field
When a charged particle, like a ball with a net charge, moves through a magnetic field, it experiences a force that can alter its path. This phenomenon is essential in understanding how particles behave in magnetic fields, such as those in the Earth's atmosphere or in man-made devices like particle accelerators.

The direction of the force is always perpendicular to both the magnetic field and the velocity of the particle. This is why in our exercise, even though the charged ball moves under the influence of gravity and the magnetic field, its motion isn't directly influenced by the magnetic field until we consider the force on it right before it hits the ground. The magnetic force doesn't do work on the particle; it only changes the direction. Hence, the horizontal speed remains constant and the gravitational force affects the vertical motion just like any other projectile.

Understanding Projectile Motion in a Magnetic Field

In the case of our charged ball, we treat its motion through the air as projectile motion and ignore the influence of the magnetic field on its trajectory until the final moment before impact. This is because the magnetic force doesn't affect the speed of the ball, only its direction, which is why we use gravitational equations to determine the time of flight.
Lorentz Force
The force exerted on a charged particle moving through a magnetic field is known as the Lorentz force. The equation for the Lorentz force is given by \( F = QvB \sin(\theta) \), where \( F \) stands for the magnetic force acting on the particle, \( Q \) is the charge of the particle, \( v \) is the velocity at which the particle is moving, \( B \) is the magnetic field's strength, and \( \theta \) is the angle between the particle's velocity and the magnetic field direction. In the case of the charged ball, the Lorentz force becomes significant when considering the magnetic force acting on it just before impact.

In the exercise, the maximum force is applied because the velocity is perpendicular to the magnetic field, resulting in \( \theta = 90^\circ \) (or \( \pi/2 \) radians), which makes \( \sin(\theta) \) equal to 1. This represents a fundamental concept in electromagnetic physics: the force is greatest when the charged particle's path is perpendicular to the field lines.
Projectile Motion Physics
Projectile motion refers to the motion of an object that is launched into the air and subject to gravity, without any significant air resistance. The object follows a curved path known as a trajectory. Typical equations of projectile motion involve calculating the time of flight, maximum height, range, and the final velocity of the object.

Our charged ball problem is an example of projectile motion with an added electromagnetic aspect. To analyze the motion, we first consider the physical principles of projectiles. Due to gravity, vertical velocity changes over time, while horizontal velocity remains constant in absence of air resistance or other horizontal forces (ignoring the magnetic force in this part of the motion). The time to reach the ground, as calculated in the exercise, is a result of the vertical component of the ball's motion.

Interplay of Gravitational and Magnetic Forces

While gravity acts continuously during the ball's flight, the magnetic force is only considered at the point just before impact in the exercise. This illustrates how different forces rule different aspects of motion: gravity governs the vertical descent, and the magnetic field interacts with the horizontal motion. In projectile motion physics, distinguishing these components is crucial for solving complex motion problems.

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Most popular questions from this chapter

Sodium ions (Nat) move at \(0.851 \mathrm{~m} / \mathrm{s}\) through a bloodstream in the arm of a person standing near a large magnet. The magnetic field has a strength of \(0.254 \mathrm{~T}\) and makes an angle of \(51.0^{\circ}\) with the motion of the sodium jons. The arm contains \(100 \mathrm{~cm}^{3}\) of blood with a concentration of \(3.00 \times 10^{20} \mathrm{Na}^{+}\) ions per cubic centimeter, If no other ions were present in the arm, what would be the magnetic force on the arm?

A lightning bolt may carry a current of \(1.00 \times 10^{4} \mathrm{~A}\) for a short time. What is the resulting magnetic field \(100 \mathrm{~m}\) from the bolt? Suppose the bolt extends far above and below the point of observation.

An electron enters a region of magnetic field of mag. nitude \(0.0100 \mathrm{~T}\), traveling perpendicular to the linear boundary of the region. The direction of the field is perpendicular to the velocity of the clectron. (a) Determine the time it takes for the electron to leave the "field-filled" region, noting that its path is a semicircle. (b) Find the kinetic energy of the electron if the radius of its semicir cular path is \(2.00 \mathrm{~cm}\).

Using an electromagnetic flowmeter (Fig. P19.69), a heart surgeon monitors the flow rate of blood through an artery. Electrodes \(A\) and \(B\) make contact with the outer surface of the blood vessel, which has interior diameter \(3.00 \mathrm{~mm} .\) (a) For a magnetic field magnitude of \(0.0400 \mathrm{~T}\), a potential difference of \(160 \mu \mathrm{V}\) appears between the electrodes. Calculate the speed of the blood. (b) Verify that electrode \(A\) is positive, as shown. Does the sign of the \(\mathrm{emf}\) depend on whether the mobile ions in the blood are predominantly positively or negatively charged? Explain.

An electron moves in a circular path perpendicular to a uniform magnetic ficld with a magnitude of \(2.0 \mathrm{mT}\). If the speed of the electron is \(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\), determine (a) the radius of the circular path and (b) the time it takes to complete one revolution.

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