Apositively charged particle of mass
7.2×10−8kg
is traveling due east with a speed of 85 m/s and enters a 0.31-t uniform magnetic field. the particle moves through one-quarter of a circle in a time of
2.2×10−3s
at which time it leaves the field heading due south. all during the motion the particle moves perpendicular to the magnetic field. (a) what is the magnitude of the magnetic force acting on the particle? (b) determine the magnitude of its charge.

(a)

(b)

Explanation:

Given:

According to the given condition the positive charge enters due east and leaves the magnetic field heading due south completing one-quarter of a circle during the course of time during which it travels perpendicular to the magnetic field.

Now, by the Fleming's left hand rule we obtain that the direction of the field is outward of the plane in which the charge has traced the locus.

Firstly we find the angular velocity:

We know,

.........................(1)

also

.......................................(2)

from eq.(1) & (2)

(b)

∵

Putting the respective values in the above expression:

(a)

Putting the respective values in eq. (2)

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

answera compound;