physics help

Question

1 - A proton travels through a magnetic field at a speed of 6.90 x 10⁵ m/s, perpendicular to the field. If the radius of curvature is 7.30 mm, what is the magnetic field strength?

2 - A particle accelerated by a potential difference enters a velocity selector. The particle travels straight when the magnetic field is 0.400 T and the electric field is 6.30 × 10⁵ V/m. Once the electric field is turned off, a sensor determines that the radius or the particle’s path is 8.19 cm. Use the charge-to-mass ratio of the particle to determine whether it is an alpha particle, electron, or proton. 

Hint: Check your data booklet for the charges and masses.

 

Question 4 options:

	alpha particle
	electron
	proton
	none of the above

 

Answer 1

(1)

The Speed of the Proton in magnetic Field is 6.90 * 10^5 m/sec

Radius of curvature is 7.30 mm = 7.3 * 10^-3 m

Mass of Proton = 1.6762 * 10^-27 kg

Charge of Proton = 1.602 * 10^-19 C

When an electron is moving in a circular path then Magnetic Force is equal to Centripetal Force

Magnetic Force = Bev

Centripetal Force = mv²/r

Bev = mv²/r

B = mv/er

B = (1.6762 * 10^-27 * 6.90 * 10^5)/(1.602 * 10^-19 * 7.3 * 10^-3)

B = 0.9889

Therefore , the magnetic field strength is 0.9889 T.

Answer 2

(b)

Electric Field is 6.30 * 10^5 v/m

Magnetic Field Strength is 0.400 T

Radius = 8.19 cm = 8.19 * 10^-2 m

When Electric Field and Magnetic Field both are acting then v = E/B

v = (6.30 * 10^5)/0.400

v = 15.75 * 10^5 m/sec

So the Velocity of the Particle is 15.75 * 10^5 m/sec

When Electric Field is turned off, Particle moves in a circular path then Magnetic Force is equal to Centripetal Force

Magnetic Force = Bqv

Centripetal Force = mv²/r

Bqv = mv²/r

q/m = v/Br

q/m = (15.75 * 10^5)/(0.400 * 8.19 * 10^-2 )

q/m = 0.480 * 10^8 C/kg

The Charge to mass ratio of Proton is in the range of 10^8.

Therefore,

The Charge to mass ratio of Proton is 0.480 * 10^8 C/kg.

The Charge to mass ratio indicates that the paticle is Proton.