A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec
tor. Ignoring the particle's weight, what type of path will the particle follow?
A) follow a spiralpath.
B) move in a straight line.C) move along a parabolic path.
D)follow a circular path.
A) follow a spiralpath.
B) move in a straight line.C) move along a parabolic path.
D)follow a circular path.
1 answer:
Answer:
The particle path will follow
(d) a circular path
Explanation:
When a charged particle having charge of magnitude '' enters into a magnetic field such that its velocity vector '
' is perpendicular to the direction of the magnetic field '
', then it will experience a force, called Lorentz force (
), given by
As shown in the figure, the magnetic field is directed perpendicular to the plane and towards the plane (as shown by the circle and 'X'-sign) and the velocity vector is from left to right on the plane.
According to the property of cross-product, the Lorentz force () acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path,as shown in the figure.
You might be interested in
Answer:
52 mm/s (approximately)
Explanation:
Given:
Initial speed of the projectile is,
Angle of projection is,
Time taken to land on the hill is,
In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.
So, the velocity in the horizontal direction always remains the same.
The horizontal component of initial velocity is given as:
Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.
Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.