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?
No
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.