Question

An ion with charge of Q = +3.2 x 10-19 C is in a region where a uniform electric field of magnitude E = 5.0 X 105 V/m is perpendicular to a uniform magnetic field B = 0.80 T. If the ion's acceleration is zero then what is its speed v? a) v = 1.6 x 10-5 m/s b) v = 4.0 x 105 m/s c) v = 6.3 x 105 m/s d) v = 0 m/s

Answer 1

Answer:

option C

Explanation:

given,

Q = +3.2 x 10⁻¹⁹ C

E = 5.0 X 10⁵ V/m

B = 0.80 T

ion's acceleration is zero  

when acceleration is zero the magnitude of both the forces becomes equal.

q E = q V B

v = $\dfrac{E}{B}$

v= $\dfrac{5 \times 10^5}{0.80}$

v = 6.25 × 10⁵ m/s ≈ 6.3 × 10⁵ m/s

hence, the correct answer is option C