A proton moves through a magnetic field at 26.1 % of the speed of light. At a location where the field has a magnitude of 0.00667 T and the proton's velocity makes an angle of 139 ∘ with the field, what is the magnitude F B of the magnetic force acting on the proton? Use c = 2.998 × 10 8 m/s for the speed of light and e = 1.602 × 10 − 19 C as the elementary charge.

Explanation:

When a charged particle is moving in a region with a magnetic field, the particle experiences a force perpendicular to its direction of motion. The magnitude of this force is given by

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

is the angle between the direction of v and B

In this problem we have:

is the charge of the proton

is the speed of the proton, where

is the speed of light

is the strength of the magnetic field

is the angle between the direction of the proton and the magnetic field

Substituting, we find the magnitude of the force:

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

the answer is b, have negligible mass