Question

C. A singly charged positive ion of mass 2.2 x 10-25 kg is moving upwards through a capacitor. The capacitor creates a constant electric field of magnitude 10,000 V/m. Inside the capacitor, the ion is not deflected due to an external magnetic field of magnitude 1.0 T that points into the page everywhere in space. i. After it exits the capacitor, is the ion going to be deflected right or left? Explain your reasoning. ii. Find the distance from where the ion exits to where it strikes the horizontal plate after turning around.

Answer 1

Step-by-step explanation:

As the particle remains un-deflected in a magnetic and electric field

velocity V = E / B , E is electric field , B is magnetic field.

V = 10000 / 1 = 10000 m /s

The particle will exit with this velocity . After that it will have circular path under Only magnetic field B .

i ) We can find the direction of force in magnetic field from left hand rule. Magnetic field is into the page and particle is going upward , then force must be towards the left . So it will exit towards the left.

ii ) The particle will have downward vertical motion after it moves through half the circular path . Distance traveled horizontally will be equal to diameter of circular path

BqV = mV² / R

R = mV / Bq

m is mass of the particle, q is charge on the particle

= 2.2 x 10⁻²⁵ x 10000 / (1 x 1.6 x 10⁻¹⁹)

= 1.375 x 10⁻² m

= 1.375 cm .

Diameter = 2R

= 2.75 cm