Answered

At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Particle [tex]\( q_1 \)[/tex] has a charge of [tex]\( 2.7 \mu C \)[/tex] and a velocity of 773 [tex]\( m/s \)[/tex]. If it experiences a magnetic force of [tex]\( 5.75 \times 10^{-3} N \)[/tex], what is the strength of the magnetic field?

In the same magnetic field, particle [tex]\( q_2 \)[/tex] has a charge of [tex]\( 42.0 \mu C \)[/tex] and a velocity of [tex]\( 1.21 \times 10^3 m/s \)[/tex]. What is the magnitude of the magnetic force exerted on particle 2?


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.