a proton moves in a circle of radius 0.4 when it enters a region with a magnetic field of 1.0t which points into the plane the s
1 answer:
Answer:
4 x 10⁷m/s
Explanation:
When a charged particle moves in a curved path in a magnetic field, it experiences some magnetic force, and in the absence of any other force, which supplies the centripetal force needed to keep the particle in balance.
Let the magnetic force be
Let the centripetal force be
=> =
--------------(i)
We know that;
= qvBsinθ
Where;
q = charge on the particle
v = speed of the particle
B = magnetic field intensity
θ = angle between the speed and magnetic field vectors
Also;
=
Where;
m = mass of the particle
v = velocity/speed of the particle
r = radius of the circular path of motion.
From equation (i)
qvBsinθ = [divide both sides by v]
qBsinθ = [make v subject of the formula]
v = qrBsinθ / m --------------------(ii)
From the question;
B = 1.0T
r = 0.4m
θ = 90° [since magnetic field is always perpendicular to velocity]
q = 1.6 x 10⁻¹⁹C [charge of a proton]
m = 1.6 x 10⁻²⁷kg [mass of a proton]
Substitute these values into equation(ii) as follows;
v = (1.6 x 10⁻¹⁹ x 0.4 x 1.0 x sin90°) / (1.6 x 10⁻²⁷)
v = 4 x 10⁷ m/s
Therefore the speed of the proton is 4 x 10⁷m/s
You might be interested in
First,
where is density,
is mass, and
is volume. We can compute the volume of the roll:
When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).
So we have
where is the given area, so
If we're taking significant digits into account, the volume we found would have been , in turn making the thickness
.
Explanation:
It is given that,
Displacement of the delivery truck, (due east)
Then the truck moves, (due south)
Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :
d = 4.03 km
Let is the direction of the truck’s displacement from the warehouse from south of east.
So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.