Due to friction with the air, an airplane has acquired a net charge of 1.70 10-5 C. The plane moves with a speed of 3.15 102 m/s at an angle with respect to the earth's magnetic field, the magnitude of which is 5.00 10-5 T. The magnetic force on the airplane has a magnitude of 2.35 10-7 N. Find the angle . (There are two possible angles.)

Question

Due to friction with the air, an airplane has acquired a net charge of 1.70 10-5 C. The plane moves with a speed of 3.15 102 m/s at an angle with respect to the earth's magnetic field, the magnitude of which is 5.00 10-5 T. The magnetic force on the airplane has a magnitude of 2.35 10-7 N. Find the angle . (There are two possible angles.)

° (smaller angle)
° (larger angle)

Answer 1

F=qv x B in vectors

so

sinTheta=F/qvB

and remember your trig identies:

sinTheta= sin(180-Theta) as I recall...

For instance, Sin 87= sin 93

Answer 2

To find the angle θ, we can use the formula for the magnetic force:

F = qvBsinθ

Where:
F = magnetic force (given as 2.35 x 10^-7 N)
q = charge of the object (given as 1.70 x 10^-5 C)
v = velocity of the object (given as 3.15 x 10^2 m/s)
B = magnitude of the magnetic field (given as 5.00 x 10^-5 T)
θ = angle between the velocity vector and the magnetic field vector (unknown)

Rearranging the equation, we can solve for sinθ:

sinθ = F / (qvB)

Let's substitute the given values:

sinθ = (2.35 x 10^-7 N) / ((1.70 x 10^-5 C) * (3.15 x 10^2 m/s) * (5.00 x 10^-5 T))

Now we can calculate sinθ:

sinθ = 0.0071

To find the angle θ, we need to take the inverse sine (sin^-1) of the calculated value:

θ = sin^-1(0.0071)

Using a calculator, we find that θ is approximately 0.408°.

Since there are two possible angles, we need to find the smaller and the larger angle. The smaller angle is simply the calculated θ (0.408°), and to find the larger angle, we subtract the calculated angle from 180°:

Larger angle = 180° - 0.408° = 179.592°

So, the two possible angles are:
Smaller angle = 0.408°
Larger angle = 179.592°

Answer 3

To find the angle θ, we can use the formula for the magnetic force on a moving charged particle:

F = q * v * B * sin(θ)

Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field strength
θ is the angle between the velocity vector and the magnetic field vector

Rearranging the formula, we have:

sin(θ) = F / (q * v * B)

Plugging in the values given:

sin(θ) = (2.35 * 10^-7 N) / ((1.70 * 10^-5 C) * (3.15 * 10^2 m/s) * (5.00 * 10^-5 T))

sin(θ) = 0.277108

Now, we can find the angle θ by taking the inverse sine (or arcsine) of the result:

θ = arcsin(0.277108)

Using a calculator, we find:

θ = 16.6107° (smaller angle)
θ = 163.389° (larger angle)

So the two possible angles are:
16.6107° (smaller angle)
163.389° (larger angle)