Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.85 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2310 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive zdirection and has a magnitude of 3.55 V/m, (b) in the negative z direction and has a magnitude of 3.55 V/m, and (c) in the positive x direction and has a magnitude of 3.55 V/m?

Answer 1

Answer 2

Given Information:

• Proton: Charge (q) = +1.602 × 10⁻¹⁹ C

• Magnetic Field (B): -x direction, magnitude = 2.85 mT = 2.85 × 10⁻³ T

• Velocity (v): +y direction, magnitude = 2310 m/s

• Electric Field (E): Varies in each part (a, b, c)

Forces Involved:

• Magnetic Force (F_B): F_B = q(v × B)

• Electric Force (F_E): F_E = qE

Calculating Magnetic Force (F_B):

• Direction: Using the right-hand rule, v (y direction) crossed with B (-x direction) gives a force in the +z direction.

• Magnitude: F_B = qvB = (1.602 × 10⁻¹⁹ C) × (2310 m/s) × (2.85 × 10⁻³ T) ≈ 1.056 × 10⁻¹⁵ N

Part (a): Electric Field (E) in +z direction, magnitude 3.55 V/m

• Electric Force (F_E):

• Direction: +z direction (same as E)

• Magnitude: F_E = qE = (1.602 × 10⁻¹⁹ C) × (3.55 V/m) ≈ 5.687 × 10⁻¹⁹ N

• Net Force (F_net):

• Direction: Both F_B and F_E are in the +z direction, so the net force is also in the +z direction.

• Magnitude: F_net = F_B + F_E = 1.056 × 10⁻¹⁵ N + 5.687 × 10⁻¹⁹ N ≈ 1.056 × 10⁻¹⁵ N (F_E is much smaller than F_B)

Part (b): Electric Field (E) in -z direction, magnitude 3.55 V/m

• Electric Force (F_E):

• Direction: -z direction (same as E)

• Magnitude: F_E = qE = (1.602 × 10⁻¹⁹ C) × (3.55 V/m) ≈ 5.687 × 10⁻¹⁹ N

• Net Force (F_net):

• Direction: F_B is in +z, F_E is in -z. They are opposite.

• Magnitude: F_net = |F_B - F_E| = |1.056 × 10⁻¹⁵ N - 5.687 × 10⁻¹⁹ N| ≈ 1.056 × 10⁻¹⁵ N (F_E is much smaller than F_B)

Part (c): Electric Field (E) in +x direction, magnitude 3.55 V/m

• Electric Force (F_E):

• Direction: +x direction (same as E)

• Magnitude: F_E = qE = (1.602 × 10⁻¹⁹ C) × (3.55 V/m) ≈ 5.687 × 10⁻¹⁹ N

• Net Force (F_net):

• Direction: F_B is in +z, F_E is in +x. They are perpendicular.

• Magnitude: F_net = √(F_B² + F_E²) = √((1.056 × 10⁻¹⁵ N)² + (5.687 × 10⁻¹⁹ N)²) ≈ 1.056 × 10⁻¹⁵ N (F_E is much smaller than F_B)

Answer 3

Given:

• Magnetic field (B): 2.85 mT (−? direction).

• Proton velocity (v): 2310 m/s (+? direction).

• Magnetic force (F_B):
  $F_B=qvB=1.055\times 10^{-18}\text{N}$ (+? direction).

• 
  

(a) Electric field (+?, 3.55 V/m):

• Electric force (F_E): $5.687\times 10^{-19}\text{N}$ (+?).

• Net force: $F_B+F_E=1.62\times 10^{-18}\text{N}$ (+?).

• 
  

(b) Electric field (−?, 3.55 V/m):

• Electric force (F_E): $5.687\times 10^{-19}\text{N}$ (−?).

• Net force: $F_B-F_E=4.86\times 10^{-19}\text{N}$ (+?).

• 
  

(c) Electric field (+?, 3.55 V/m):

• Electric force (F_E): $5.687\times 10^{-19}\text{N}$ (+?).

• Net force: Combines perpendicular forces:
  $\sqrt{F_E^2+F_B^2}=1.20\times 10^{-18}\text{N}$ (diagonal in ?-? plane).

• 
  

Summary:-

• (a) Strongest net force in +?.

• (b) Weaker net force in +?.

• (c) Net force at an angle, combining ? and ?