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Chapter 27: Problem 1

A particle with a charge of -1.24 \(\times\) 10\(^{-8}\) C is moving with instantaneous velocity \(\vec{v} =\) 14.19 \(\times\) 10\(^4\) m/s)\(\hat{\imath}\) + (-3.85 \(\times\) 10\(^4\) m/s)\(\hat{\jmath}\). What is the force exerted on this particle by a magnetic field (a) \(\overrightarrow{B} =\) (1.40 T)\(\hat{\imath}\) and (b) \(\overrightarrow{B} =\) (1.40 T) \(\hat{k}\) ?

Short Answer

Expert verified
(a) \( \vec{F_1} = 0 \hat{\imath} + 0 \hat{\jmath} - 6.68 \times 10^{-4} \hat{k} \text{ N} \); (b) \( \vec{F_2} = 6.68 \times 10^{-4} \hat{\imath} + 2.47 \times 10^{-3} \hat{\jmath} \text{ N} \).

Step by step solution

01

Understand the Given Problem

We need to find the force exerted by a magnetic field on a charge with a specific velocity vector in two different scenarios. The charge is given as \( q = -1.24 \times 10^{-8} \text{ C} \), and the velocity vector is \( \vec{v} = 14.19 \times 10^4 \hat{\imath} - 3.85 \times 10^4 \hat{\jmath} \text{ m/s} \). We need to find the force for two different magnetic fields: \( \overrightarrow{B_1} = 1.40 \hat{\imath} \text{ T} \) and \( \overrightarrow{B_2} = 1.40 \hat{k} \text{ T} \).
02

Use the Lorentz Force Formula

The magnetic force \( \vec{F} \) on a charged particle moving with velocity \( \vec{v} \) in a magnetic field \( \overrightarrow{B} \) is given by the cross product \( \vec{F} = q(\vec{v} \times \overrightarrow{B}) \). We'll need to compute this for each magnetic field scenario.
03

Calculate Cross Product for (a) \( \overrightarrow{B} = 1.40 \hat{\imath} \)

For \( \overrightarrow{B_1} = 1.40 \hat{\imath} \), the cross product is \( \vec{v} \times \overrightarrow{B_1} = (14.19 \times 10^4 \hat{\imath} - 3.85 \times 10^4 \hat{\jmath}) \times 1.40 \hat{\imath} \). Since a vector crossed with itself is zero, and the cross product between vectors on the same axis is zero, this simplifies to: \( \vec{v} \times \overrightarrow{B_1} = 0 \hat{\imath} + 0 \hat{\jmath} + (-1.40 \times (-3.85 \times 10^4)) \hat{k} \). Calculating gives: \( \vec{v} \times \overrightarrow{B_1} = 0 \hat{\imath} + 0 \hat{\jmath} + 5.39 \times 10^4 \hat{k} \).
04

Calculate Force for (a)

Using \( q = -1.24 \times 10^{-8} \text{ C} \) and the cross product from Step 3, we find: \( \vec{F_1} = -1.24 \times 10^{-8} \times (0 \hat{\imath} + 0 \hat{\jmath} + 5.39 \times 10^4 \hat{k}) \). This gives: \( \vec{F_1} = 0 \hat{\imath} + 0 \hat{\jmath} - 6.68 \times 10^{-4} \hat{k} \text{ N} \).
05

Calculate Cross Product for (b) \( \overrightarrow{B} = 1.40 \hat{k} \)

For \( \overrightarrow{B_2} = 1.40 \hat{k} \), the cross product is \( \vec{v} \times \overrightarrow{B_2} = (14.19 \times 10^4 \hat{\imath} - 3.85 \times 10^4 \hat{\jmath}) \times 1.40 \hat{k} \).The result is calculated using the determinant method for cross products:\( ( -3.85 \times 10^4 \times 1.40) \hat{\imath} - (14.19 \times 10^4 \times 1.40) \hat{\jmath} \). Which results in:\( \vec{v} \times \overrightarrow{B_2} = -5.39 \times 10^4 \hat{\imath} - 1.99 \times 10^5 \hat{\jmath} + 0 \hat{k} \).
06

Calculate Force for (b)

Using \( q = -1.24 \times 10^{-8} \text{ C} \) and the cross product from Step 5, we find:\( \vec{F_2} = -1.24 \times 10^{-8} (-5.39 \times 10^4 \hat{\imath} - 1.99 \times 10^5 \hat{\jmath}) \).This results in: \( \vec{F_2} = 6.68 \times 10^{-4} \hat{\imath} + 2.47 \times 10^{-3} \hat{\jmath} \text{ N} \).

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Key Concepts

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

Magnetic Field
A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. It's visualized as "lines of force" that exit through a magnet's north pole and enter through its south pole. Magnetic fields can be created by permanent magnets or flows of electric current. When a moving charge enters a magnetic field, it experiences a force. This force's direction depends on the charge's velocity and the magnetic field's direction. The strength of a magnetic field is measured in Tesla (T).

Key Points:
  • A magnetic field exerts a force on moving charges.
  • The magnetic field's direction and the charge鈥檚 motion will determine the direction of this force.
  • Strong magnetic fields exert stronger forces on charges.
In the context of the original exercise, recognizing which direction the magnetic fields (\( \overrightarrow{B_1} \) and \( \overrightarrow{B_2} \)) are acting in is crucial for calculating the resulting force on the charged particle.
Cross Product
The cross product is a vector multiplication operation that yields a vector perpendicular to two input vectors in three-dimensional space. This operation is central in physics, particularly in electromagnetism, where it is used to determine the force experienced by a charged particle within a magnetic field. Mathematically, the cross product of two vectors \( \vec{a} \) and \( \vec{b} \) is denoted as \( \vec{a} \times \vec{b} \). The magnitude of the resulting vector is given by \(|\vec{a}| |\vec{b}| \sin \theta\), where \( \theta \) is the angle between the original vectors.
Key Characteristics:
  • A cross product of parallel vectors is zero since \( \sin(0^\circ) = 0 \).
  • The resulting vector is perpendicular to the original vectors.
  • It is used to calculate torques, rotations, and forces like the Lorentz force.
In the exercise, the cross product helps determine \( \vec{v} \times \overrightarrow{B} \), which then allows calculation of the Lorentz force on the particle through \( \vec{F} = q(\vec{v} \times \overrightarrow{B}) \).
Vector Calculus
Vector calculus is a mathematical tool used to study vector fields, which are functions assigning a vector to each point in space. It's vital in physics for understanding phenomena involving fields, such as electromagnetic fields. Vector operations, such as addition, scalar multiplication, and cross products, help solve complex physical problems.

Important Operations:
  • Vector Addition: Combines two vectors' magnitudes and directions.
  • Scalar Multiplication: Involves multiplying a vector by a scalar (number), scaling its magnitude.
  • Cross Product: Yields a vector perpendicular to two given vectors, especially crucial in calculating forces in electromagnetism.
In our task, vector calculus is applied to the velocity vector \( \vec{v} \) and the magnetic field \( \overrightarrow{B} \) to determine the force using the cross product. This application exemplifies the real-world utility of vector calculus in physics.

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Most popular questions from this chapter

A conducting bar with mass m and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{B}\) fills the region between the rails (\(\textbf{Fig. P27.65}\)). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\), find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth (11.2 km/s). Let \(B =\) 0.80 T, \(I =\) 2.0 \(\times\) 10\(^3\) A, \(m =\) 25 kg, and \(L =\) 50 cm. For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A singly charged ion of \(^7\)Li (an isotope of lithium) has a mass of 1.16 \(\times\) 10\(^{-26}\) kg. It is accelerated through a potential difference of 220 V and then enters a magnetic field with magnitude 0.874 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 \(\Omega\). When the motor is running at full load on a 120-V line, the emf in the rotor is 105 V. (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

The amount of meat in prehistoric diets can be determined by measuring the ratio of the isotopes \(^{15}\)N to \(^{14}\)N in bone from human remains. Carnivores concentrate \(^{15}\)N, so this ratio tells archaeologists how much meat was consumed. For a mass spectrometer that has a path radius of 12.5 cm for \(^{12}\)C ions (mass 1.99 \(\times\) 10\(^{-26}\) kg), find the separation of the \(^{14}\)N 1mass 2.32 \(\times\) 10\(^{-26}\) kg2 and 15N (mass 2.49 \(\times\) 10\(^{-26}\) kg) isotopes at the detector.

A wire 25.0 cm long lies along the \(z\)-axis and carries a current of 7.40 A in the \(+z\)-direction. The magnetic field is uniform and has components \(B_x =\) -0.242 T, \(B_y =\) -0.985 T, and \(B_z\) = -0.336 T. (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the net magnetic force on the wire?
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