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QuestionA proton is fired with a speed of 0.90×106 m/s0.90 \times 10^{6} \mathrm{~m} / \mathrm{s} through the parallel-plate capacitor shown in (Figure 1). The capacitor's electric field is E⃗=(0.50×105 V/m, down )\vec{E}=\left(0.50 \times 10^{5} \mathrm{~V} / \mathrm{m}, \text { down }\right)
Part A
What is the strength of the magnetic field B⃗\vec{B} that must be applied to allow the proton to pass through the capacitor with no change in speed or direction? Express your answer with the appropriate units.

Studdy Solution

STEP 1

1. The proton is moving with a constant velocity through the capacitor.
2. The electric field is uniform and directed downward.
3. The magnetic field must counteract the electric field to prevent deflection.
4. The proton's speed and direction remain unchanged.

STEP 2

1. Recall the force balance condition for a charged particle moving in electric and magnetic fields.
2. Set up the equation for the forces acting on the proton.
3. Solve for the magnetic field strength.

STEP 3

Recall the force balance condition for a charged particle moving in electric and magnetic fields:
The electric force F⃗E \vec{F}_E and magnetic force F⃗B \vec{F}_B must cancel each other out for the proton to pass through without deflection. This gives:
F⃗E+F⃗B=0 \vec{F}_E + \vec{F}_B = 0

STEP 4

Set up the equation for the forces acting on the proton:
The electric force on the proton is given by:
F⃗E=qE⃗ \vec{F}_E = q \vec{E}
The magnetic force on the proton is given by:
F⃗B=qv⃗×B⃗ \vec{F}_B = q \vec{v} \times \vec{B}
For the forces to cancel:
qE⃗+qv⃗×B⃗=0 q \vec{E} + q \vec{v} \times \vec{B} = 0
This simplifies to:
E⃗=−v⃗×B⃗ \vec{E} = -\vec{v} \times \vec{B}

STEP 5

Solve for the magnetic field strength:
Given that the proton's velocity v⃗ \vec{v} is perpendicular to the magnetic field B⃗ \vec{B} , the magnitude of the magnetic force is:
E=vB E = vB
Substitute the given values:
0.50×105 V/m=0.90×106 m/s×B 0.50 \times 10^{5} \, \text{V/m} = 0.90 \times 10^{6} \, \text{m/s} \times B
Solve for B B :
B=0.50×105 V/m0.90×106 m/s B = \frac{0.50 \times 10^{5} \, \text{V/m}}{0.90 \times 10^{6} \, \text{m/s}}
B=0.500.90×10−1 T B = \frac{0.50}{0.90} \times 10^{-1} \, \text{T}
B≈0.556×10−1 T B \approx 0.556 \times 10^{-1} \, \text{T}
B≈0.056 T B \approx 0.056 \, \text{T}
The strength of the magnetic field required is:
0.056 T \boxed{0.056 \, \text{T}}

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