Question

At \(t=0\) an electron crosses the positive \(y\) -axis \((\) so \(x=0)\) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip from \(y\) -axis to \(x\) -axis take?

Short answer

Answer: The magnitude and direction of the magnetic field is \(1.885 \times 10^{-4} T\) in the negative z-direction. 2. How much work is done on the electron during its motion? Answer: The work done on the electron during this motion is 0. 3. How much time does it take for the electron to move from the y-axis to the x-axis? Answer: The time taken for the trip from the y-axis to the x-axis is \(0.94 \times 10^{-8} s\).

Step-by-step solution

Identify the given information

We are given the following information: - Initial position of the electron: \((0, 60.0~cm)\) - Initial velocity of the electron: \(2.00 \times 10^5~m/s\) in the positive x-direction - Uniform magnetic field acting on the electron. We are asked to find: a) The magnitude and direction of the magnetic field b) The work done on the electron c) The time taken for the motion

Find the magnetic field

Since the electron is moving in a straight line in the x-direction, the magnetic force on the electron would be perpendicular to the velocity vector. We know, \(F = q(v \times B)\), where F is the magnetic force, q is the charge of the electron, v is the velocity, and B is the magnetic field. In this case, the magnetic force acts as the centripetal force causing the electron to take a circular path. Thus, we have: \(F = \dfrac{mv^2}{r}\), where m is the mass of the electron and r is the radius of the circular path. Since the electron crosses the x-axis at \((60.0~cm,0)\), the radius of the circle would be 60.0 cm: Now, equating the two forces, we get: \(q(v \times B) = \dfrac{mv^2}{60.0 \times 10^{-2} m}\) Let's solve for B: \(B = \dfrac{m}{q} \cdot \dfrac{v}{60.0 \times 10^{-2} m}\) Using the electron's mass \(m = 9.11 \times 10^{-31} kg\) and charge \(q = 1.6 \times 10^{-19} C\), we get: \(B = \dfrac{9.11 \times 10^{-31} kg}{1.6 \times 10^{-19} C} \cdot \dfrac{2.00 \times 10^5 m/s}{60.0 \times 10^{-2} m}\) \(B = 1.885 \times 10^{-4} T\) The magnetic field will be in the negative z-direction as the force on the electron is upwards to make it cross the x-axis. Hence, the magnitude and direction of B is \(1.885 \times 10^{-4} T\) in the negative z-direction.

Calculate the work done on the electron

The work done on the electron by the magnetic force is given by: \(W = F \cdot d \cdot \cos \theta\) Since the magnetic force is perpendicular to the direction of displacement (path of the electron), \(\theta = 90°\) and \(\cos \theta = 0\): \(W = F \cdot d \cdot 0 = 0\) Thus, no work is done on the electron during this motion.

Find the time taken for the trip

To find the time taken for the motion, we determine the time period of an electron undergoing a circular motion under the influence of the magnetic field. The time period, T, is given by: \(T = \dfrac{2 \pi r}{v}\) Using the given values for r and v: \(T = \dfrac{2 \pi (60.0 \times 10^{-2} m)}{2.00 \times 10^5 m/s}\) \(T = 1.88 \times 10^{-8} s\) Therefore, the trip from the y-axis to the x-axis will take half the time period: \(t = \dfrac{T}{2} = 0.94 \times 10^{-8} s\) #Answer#: a) The magnitude and direction of the magnetic field is \(1.885 \times 10^{-4} T\) in the negative z-direction. b) The work done on the electron during this motion is 0. c) The time taken for the trip from the y-axis to the x-axis is \(0.94 \times 10^{-8} s\).

Key concepts

Lorentz Force

The Lorentz force is a fundamental concept in physics which describes the force experienced by a charged particle moving through a magnetic field. Mathematically, it's expressed as \( F = q(v \times B) \) where \( F \) is the magnetic force, \( q \) is the electric charge of the particle, \( v \) is its velocity vector, and \( B \) is the magnetic field vector. When these vectors are not parallel, their cross product is nonzero, resulting in a force.
In the case of an electron (like in our exercise), which is negatively charged and moving in a magnetic field, this force will be perpendicular to both the velocity of the electron and the magnetic field lines. This perpendicular nature of the Lorentz force is what causes the electron to follow a curved path rather than a straight line. Understanding how this force dictates the trajectory of electrons is critical not only for this problem but also for broader applications in fields such as electromagnetism and particle physics.

Centripetal Force in Magnetic Field

The centripetal force is the inward force that is necessary for an object to follow a circular path. It's always directed towards the center of the rotation. When an electron travels through a magnetic field, the Lorentz force acts as this centripetal force. The relationship between the magnetic force and the centripetal force for a particle with mass \( m \) moving with speed \( v \) along a path of radius \( r \) is given by \( F = \frac{mv^2}{r} \).
Applying this to our exercise, the electron follows a half-circle from the positive y-axis to the positive x-axis which signifies that the magnetic force is providing the necessary centripetal force to sustain this circular motion. Computing the magnetic field requires equating this centripetal force to the magnetic Lorentz force exerted on the electron.

Work Done by Magnetic Force

An interesting physical concept comes into play concerning the work done by the magnetic force on a charged particle. Work is generally calculated by the equation \( W = F \cdot d \cdot \cos \theta \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement vectors. Since the magnetic force always acts perpendicular to the direction of motion, the angle is 90 degrees, leading to \( \cos(90°) = 0 \).
This means that, although the magnetic force is responsible for changing the direction of the velocity of the electron, it does not do 'work' on it since it does not change the kinetic energy of the electron. This is a crucial insight students should gain from solving this problem. It is equally important in understanding that in a magnetic field, charged particles exchange directional momentum without exchanging energy.

Period of Electron in Magnetic Field

The period of an electron in a magnetic field is the time it takes for the electron to complete one full circular cycle. It's given by the formula \( T = \frac{2\pi r}{v} \), where \( r \) is the radius of the circular path, and \( v \) is the speed of the electron. This is derived from the concept of circular motion, where the period is the distance traveled in one complete circle (the circumference which is \( 2\pi r \) ) divided by the constant speed of the electron.
For our electron traveling from the y-axis to the x-axis across a uniform magnetic field, we are dealing with a semi-circular path. Thus, to find the time it takes to cross this path, we use half the period, which results from the symmetrical nature of the motion. It's crucial for students to understand that the magnetic field's strength and the electron's speed dictate this period, highlighting the predictability of motion under electromagnetic forces.