A charge of -5.02 μC is traveling at a speed of 8.86 × 106 m/s in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is 54.6o. A force of magnitude7.85 × 10-3 N acts on the charge. What is the magnitude of the magnetic field?
5 years ago
11 months ago
To find the magnitude of the magnetic field, we can use the equation for the magnetic force on a charged particle:
F = q * v * B * sin(θ)
Where:
F is the force on the charge (given as 7.85 × 10^-3 N)
q is the charge of the particle (-5.02 μC or -5.02 × 10^-6 C)
v is the velocity of the particle (8.86 × 10^6 m/s)
B is the magnitude of the magnetic field we want to find
θ is the angle between the velocity of the charge and the magnetic field (54.6°)
Now we can rearrange the equation to solve for B:
B = F / (q * v * sin(θ))
Plugging in the given values:
B = (7.85 × 10^-3 N) / ((-5.02 × 10^-6 C) * (8.86 × 10^6 m/s) * sin(54.6°))
First, evaluate the sine of 54.6°:
sin(54.6°) ≈ 0.8
Now, substitute the values into the equation:
B ≈ (7.85 × 10^-3 N) / ((-5.02 × 10^-6 C) * (8.86 × 10^6 m/s) * 0.8)
Finally, calculate the magnitude of the magnetic field:
B ≈ 0.176 T
Therefore, the magnitude of the magnetic field is approximately 0.176 Tesla.