A charge of 12 ìC, traveling with a speed of 9.0 x 106 m/s in a direction perpendicular to a magnetic field, experiences a magnetic force of 8.7 x 10-3 N. What is the magnitude of the field?
To find the magnitude of the magnetic field, we can use the formula for the magnetic force on a charged particle moving in a magnetic field:
F = q * v * B
Where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field.
We can rearrange the formula to solve for B:
B = F / (q * v)
Given:
q = 12 μC = 12 x 10-6 C
v = 9.0 x 106 m/s
F = 8.7 x 10-3 N
Plugging in these values, we can calculate the magnitude of the magnetic field:
B = (8.7 x 10-3 N) / (12 x 10-6 C * 9.0 x 106 m/s)
B ≈ 6.08 T (Tesla)
Therefore, the magnitude of the magnetic field is approximately 6.08 Tesla.
To find the magnitude of the magnetic field, we can use the formula for the magnetic force on a moving charge:
F = q * v * B
Where:
F is the magnetic force (given as 8.7 x 10^-3 N)
q is the charge (given as 12 μC, which we need to convert to coulombs: 12 μC = 12 x 10^-6 C)
v is the speed of the charge (given as 9.0 x 10^6 m/s)
B is the magnetic field (what we need to find)
Rearranging the formula, we can solve for B:
B = F / (q * v)
Now we can substitute the given values into the formula and calculate the magnitude of the magnetic field:
B = (8.7 x 10^-3 N) / ((12 x 10^-6 C) * (9.0 x 10^6 m/s))
First, multiply the charge and the speed:
B = (8.7 x 10^-3 N) / (108 x 10^-12 C * m/s)
Next, divide the force by the product of charge and speed:
B = 8.7 x 10^-3 N / 1.08 x 10^-5 C * m/s
Finally, divide the numerical values and keep the scientific notation:
B ≈ 0.80555555556 x 10^-3 T
So, the magnitude of the magnetic field is approximately 0.805 x 10^-3 T or 8.05 x 10^-4 T.