Problem 58 - University Physics with Modern Physics

Question

emf in a Bullet. At the equator, the earth's magnetic field is approximately horizontal, is directed towand the north, and has a value of \(8 \times 10^{-5} \mathrm{T}\) . (a) Estimate the emf induced between the top and bottom of a bullet shot horizontally at a target on the equator if the bullet is shot toward the east. Assume the bullet has a length of 1 \(\mathrm{cm}\) and a diamcter of 0.4 \(\mathrm{cm}\) and is traveling at 300 \(\mathrm{m} / \mathrm{s}\) . Which is at higher potential: the top or boutom of the bullet? (b) What is the emfif the bullet travels south?(c) What is the emf induced between the front and back of the bullet for any horizontal velocity?

Step-by-Step Solution

Answer

(a) 9.6 μV with top at higher potential, (b) 0 V, (c) 0 V.

Step 1: Analyze the Situation

Employ the formula for the electromotive force (emf) induced due to motion in a magnetic field. The emf (voltage) can be calculated using the relationship \( \text{emf} = B \times v \times L \), where \( B \) is the magnetic field strength, \( v \) is the velocity of the object, and \( L \) is the length of the object perpendicular to the velocity and magnetic field. Here, \( B = 8 \times 10^{-5} \ \text{T} \), \( v = 300 \ \text{m/s} \), and \( L = 0.004 \ \text{m} \) since the induced emf is perpendicular to both the magnetic field and the velocity.

Step 2: Calculate the Induced EMF for Eastward Motion

Substitute the given values into the formula:\[\text{emf} = B \times v \times L = (8 \times 10^{-5} \ \text{T}) \times (300 \ \text{m/s}) \times (0.004 \ \text{m})\]This calculation results in:\[\text{emf} = 9.6 \times 10^{-5} \ \text{V}\]When the bullet is shot east, based on Earth's magnetic field direction (northward), the top of the bullet will be at a higher potential than the bottom.

Step 3: Consider Southward Motion

If the bullet is shot toward the south, the length of the bullet is parallel to the magnetic field lines. The angle between the velocity vector and magnetic field is therefore zero, resulting in no perpendicular component, and thus:\(\text{emf} = 0 \ \text{V}\)

Step 4: Check Front-to-Back EMF for Any Horizontal Velocity

For any horizontal velocity, the length between the front and back of the bullet is parallel to the bullet's velocity, resulting in no perpendicular component to induce an emf. Thus, regardless of the direction (eastward, westward, or southward):\(\text{emf} = 0 \ \text{V}\)

Key Concepts

Electromotive Force (EMF)

To understand how electromotive force (EMF) is generated, think of it as the voltage induced across an object when it moves through a magnetic field. This happens because, in motion, the charged particles inside the object experience a force due to the magnetic field. The formula used to calculate EMF in these situations is \[\text{emf} = B \times v \times L\]where:

• \( B \) is the magnetic field strength.
• \( v \) is the velocity of the object.
• \( L \) is the length of the object perpendicular to both velocity and magnetic field.

When a bullet is fired through a horizontal magnetic field, the northward direction of Earth’s field at the equator provides a perfect example of how EMF is induced. As the bullet travels east, the upper part is at a higher potential compared to the lower part due to the direction of Earth's magnetic field being perpendicular to the bullet’s path.
If the bullet travels south, there is no EMF generated because the motion is parallel to the field lines, leading to no perpendicular component. Similarly, no EMF is induced between the front and back because the movement is aligned with the bullet's axis.

Magnetic Field

The Earth's magnetic field is a natural, invisible force field that extends around the planet. It's similar to the field generated by a bar magnet, with its magnetic field lines running from the south to the north magnetic pole. The magnetic field at the equator is mostly horizontal to the Earth’s surface and is directed toward the north. The unit used to measure magnetic field strength is the Tesla (T).

In the exercise, the Earth's magnetic field is specified as approximately \(8 \times 10^{-5} \ \text{T}\). This field affects any moving projectile, such as a bullet, by inducing an EMF due to the interaction between the field and the projectile's velocity. Importantly, the strength and direction of this magnetic field play crucial roles in determining where and how much EMF is generated.

Understanding magnetic fields helps explain how electricity and magnetism are interconnected. Applications of these principles are seen in electric generators where mechanical energy is converted to electrical energy using rotational motion within magnetic fields.

Projectile Motion

Projectile motion refers to the motion of an object thrown or projected into the air under the influence of gravity, following a curved path called a parabola. In the context of this exercise, we're dealing with a bullet, which is a classic example of a projectile moving horizontally after being fired.

When analyzing projectile motion, it's important to consider variables such as initial velocity, angle of projection, and forces acting on the projectile. In our exercise, the bullet is moving at a velocity of \(300 \ \text{m/s}\) perpendicular to the Earth's magnetic field. This fast, horizontal motion emphasizes the role of the magnetic field in generating EMF.

Regardless of the direction (east, south, or any horizontal path), the projectile’s velocity interacts differently with the magnetic field. Thus, understanding the relationship between motion and field interaction is indispensable in comprehending how various forces act on projectiles with different paths and speeds.