Question

electron is movins Vhat is the magnitude of the force acting on the electron? dicular to a 1.OT magnetic field. The electron has a (A) 1.6 x 10-13N (B)1.6x 10"2 N (C) 1.6x 10-10N (D) 1.6 x 10 N 2. The Points A through E on the Figurel represent various distances from the center of the current-carrying wire. Which point experiences the strongest magnetic field? ( (A) Point A (B) Point C (C) Point D (D) Point E 3.A wave has a frequency of 100 Hz and a wavelength of 1 m. What is the velocity of the wave? (A) 0.01 m/s (B) 1 m/s (C) 10 m/s (D) 100 m/s Figurel 4. A wave of light (- within the material is 510 nm. What is the index of refraction o 638 nm) enters a block of material. The wavelength of light f the material? ) (A)0.5 (B) 0.8 (C) 1.2 (D) 125

Answer 1

1. Magnitude of the force acting on an electron which will be given by -

F = q v B sin $\theta$

where, q = charge on an electron = 1.6 x 10^-19 C

v = velocity of an electron = 1 x 10⁷ m/s

B = magnetic field = 1 T

then, we get

F = [(1.6 x 10^-19 C) (1 x 10⁷ m/s) (1 T)] sin 90⁰

F = 1.6 x 10^-12 N

Option (B) : it will be correct.

3. The velocity of a wave which will be given by -

v = $\lambda$ / f

where, $\lambda$ = wavelength of a wave = 1 m

f = frequency of a wave = 100 Hz

then, we get

v = [(1 m) / (100 Hz)]

v = 0.01 m/s

Option (A) : it will be correct.

4. The index of refraction of a material which will be given by -

n = $\lambda$ / $\lambda$_n

where, $\lambda$ = wavelength of a wave = 638 nm

$\lambda$_n = wavelength of light within a material = 510 nm

then, we get

n = [(638 nm) / (510 nm)]

n = 1.25

Option (D) : it will be correct.