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Sagot :
To solve this problem, we need to follow two main steps:
1. Determine the strength of the magnetic field, [tex]\( B \)[/tex].
2. Use this magnetic field strength to find the magnetic force on the second particle.
### Step 1: Calculate the Magnetic Field Strength [tex]\( B \)[/tex]
Given for particle [tex]\( q_1 \)[/tex]:
- Charge, [tex]\( q_1 = 2.7 \mu C = 2.7 \times 10^{-6} \)[/tex] C
- Velocity, [tex]\( v_1 = 773 \)[/tex] m/s
- Magnetic force, [tex]\( F_1 = 5.75 \times 10^{-3} \)[/tex] N
The magnetic force on a particle moving perpendicular to a magnetic field is given by:
[tex]\[ F = q \cdot v \cdot B \][/tex]
Rearranging to solve for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{F_1}{q_1 \cdot v_1} \][/tex]
Substituting the known values:
[tex]\[ B = \frac{5.75 \times 10^{-3}}{2.7 \times 10^{-6} \cdot 773} \][/tex]
After simplifying:
[tex]\[ B \approx 2.755 \, T \][/tex]
Thus, the strength of the magnetic field [tex]\( B \)[/tex] is approximately [tex]\( 2.755 \)[/tex] Tesla.
### Step 2: Calculate the Magnetic Force on Particle [tex]\( q_2 \)[/tex]
Given for particle [tex]\( q_2 \)[/tex]:
- Charge, [tex]\( q_2 = 42.0 \mu C = 42.0 \times 10^{-6} \)[/tex] C
- Velocity, [tex]\( v_2 = 1.21 \times 10^3 \)[/tex] m/s
Using the magnetic field strength we found, [tex]\( B = 2.755 \)[/tex] T, we can now find the magnetic force on [tex]\( q_2 \)[/tex] using the formula:
[tex]\[ F_2 = q_2 \cdot v_2 \cdot B \][/tex]
Substituting the known values:
[tex]\[ F_2 = (42.0 \times 10^{-6}) \cdot (1.21 \times 10^3) \cdot 2.755 \][/tex]
After simplifying:
[tex]\[ F_2 \approx 0.14 \, N \][/tex]
Thus, the magnetic force exerted on particle [tex]\( q_2 \)[/tex] is approximately [tex]\( 0.14 \)[/tex] Newtons.
1. Determine the strength of the magnetic field, [tex]\( B \)[/tex].
2. Use this magnetic field strength to find the magnetic force on the second particle.
### Step 1: Calculate the Magnetic Field Strength [tex]\( B \)[/tex]
Given for particle [tex]\( q_1 \)[/tex]:
- Charge, [tex]\( q_1 = 2.7 \mu C = 2.7 \times 10^{-6} \)[/tex] C
- Velocity, [tex]\( v_1 = 773 \)[/tex] m/s
- Magnetic force, [tex]\( F_1 = 5.75 \times 10^{-3} \)[/tex] N
The magnetic force on a particle moving perpendicular to a magnetic field is given by:
[tex]\[ F = q \cdot v \cdot B \][/tex]
Rearranging to solve for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{F_1}{q_1 \cdot v_1} \][/tex]
Substituting the known values:
[tex]\[ B = \frac{5.75 \times 10^{-3}}{2.7 \times 10^{-6} \cdot 773} \][/tex]
After simplifying:
[tex]\[ B \approx 2.755 \, T \][/tex]
Thus, the strength of the magnetic field [tex]\( B \)[/tex] is approximately [tex]\( 2.755 \)[/tex] Tesla.
### Step 2: Calculate the Magnetic Force on Particle [tex]\( q_2 \)[/tex]
Given for particle [tex]\( q_2 \)[/tex]:
- Charge, [tex]\( q_2 = 42.0 \mu C = 42.0 \times 10^{-6} \)[/tex] C
- Velocity, [tex]\( v_2 = 1.21 \times 10^3 \)[/tex] m/s
Using the magnetic field strength we found, [tex]\( B = 2.755 \)[/tex] T, we can now find the magnetic force on [tex]\( q_2 \)[/tex] using the formula:
[tex]\[ F_2 = q_2 \cdot v_2 \cdot B \][/tex]
Substituting the known values:
[tex]\[ F_2 = (42.0 \times 10^{-6}) \cdot (1.21 \times 10^3) \cdot 2.755 \][/tex]
After simplifying:
[tex]\[ F_2 \approx 0.14 \, N \][/tex]
Thus, the magnetic force exerted on particle [tex]\( q_2 \)[/tex] is approximately [tex]\( 0.14 \)[/tex] Newtons.
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