An electron is launched with velocityv in a uniformmagnetic fieldB . The angle betweenv andB liesbetween 0 and2 . Its velocity vectorv returns to itsinitial value in a time interval of
Question
An electron is launched with velocityv in a uniformmagnetic fieldB . The angle betweenv andB liesbetween 0 and2 . Its velocity vectorv returns to itsinitial value in a time interval of
Solution
The electron moves in a circular path due to the magnetic field. The time it takes for the electron to complete one full circle and return to its initial velocity vector is called the period (T) of the motion.
The force on the electron due to the magnetic field is given by F = qvBsinθ, where q is the charge of the electron, v is its velocity, B is the magnetic field strength, and θ is the angle between v and B. This force provides the centripetal force for the circular motion, which is mv^2/r, where m is the mass of the electron and r is the radius of the circle.
Setting these two expressions for the force equal to each other gives qvBsinθ = mv^2/r. Solving for r gives r = mv/(qBsinθ).
The circumference of the circle is 2πr, so the time to complete one full circle is the circumference divided by the speed, which gives T = 2πr/v = 2πm/(qBsinθ).
So, the time interval for the velocity vector v to return to its initial value is T = 2πm/(qBsinθ).
Similar Questions
An electron moves in circular motion in a uniform magnetic field.P velocityThe velocity of the electron at point P is 6.8 × 105 m s-1 in the direction shown.The magnitude of the magnetic field is 8.5 T.(a) State the direction of the magnetic field.
An electron at point in figure has a speed of v0 = 1.41 x 106 m/s.Finda) the magnitude and direction of the magnetic field that willcause the electron to follow the semicircular path from A to Bandb) the time required for the electron to move from A to B.
If an electron is travelling horizontally towards east. A magnetic field in vertically downward direction exerts a force on the electron along*1 pointeastwestnorthsouthIf a charged particle is moving along a magnetic field line. The magnetic force on the particle is*1 pointalong its velocityopposite to its velocityperpendicular to its velocityZeroA magnetic field exerts no force on*1 pointa stationary electric chargea magnetan electric charge moving perpendicular to its directiona non-magnetized iron barAn electron beam enters a magnetic field at right angles to it as shown in the figure. The direction of force acting on the electron beam will be*1 pointTo the leftto the rightinto the planeout of the plane
An electron moves in the direction of a uniform electric field. Its initial speed is5.00 × 106 m s−1 and stops momentarily after moving through a distance of 20.0 cm.(a) What is the strength of the electric field?(b) Why the electron stops momentarily?
An electron moving at right angles to a 0.10-T magnetic field experiences an acceleration of 6.0x10'5 m/s. (a) What's its speed? (b) By how much does its speed change in 1 ns?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.