Question

In the figure, a small particle of charge = -2.0 × ${{10}}^{{-6}}$ C and mass kg has velocity m/s as it enters a region of uniform magnetic field. Theparticle is observedto travel in the semicircular path shown, withradius = 5.0 cm.

maginitude of the magnetic field is0.24 T

magnetic field directed:
to the right.
to the left
into the screen
out of the screen

Answer 1

the centripetal force acting on the particle isF = (1/2) * (m * D_o^2/R) ----------(1)the magnetic force acting on the particle isF = q * v * B * sinθ ----------(2)from (1) and (2)q *D_o * B * sinθ = (1/2) * (m *D_o^2/R)or B = (1/2) * (m * D_o/R) * (1/q * sinθ)m = 3.11 * 10^-12 kgD_o= 7.8 * 10^3 m/sR = 5.0 cm = 5.0 * 10^-2 mq = -2.0 * 10^-6 Cθ = 90^{oIn the begining of the motion of the charge,the charge ismoving in the upward direction therefore the current associatedwith the moving chargeis in upward direction.According to Ampere'sright hand thumb rule,when the direction of the current in the wireis indicated by the thumb of theright hand then the curling offingers indicates the direction of magnetic field.Here,the currentis flowing in the upward direction and hence themagnetic fieldshould be in the outward direction that is the magnetic field isout of the screen.}