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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?

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

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.

User Icon for Explain Bot Explain Bot answered
11 months ago

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.