A proton, traveling with a velocity of 6.8 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 8.6 × 10-14 N and direction of due south. What are the magnitude and direction of the magnetic field causing the force? If the field is up, then enter a number greater than zero. If the field is down, then enter a number less than zero
Bqv=6.8 × 106*8.6 × 10-14 *1.6E-19
is the answer in T? I had 9.3568E-26 and it says the answer is wrong
you are looking for B, so the equation would be
B= F/qv
To find the magnitude and direction of the magnetic field causing the force acting on the proton, we can use the formula for the magnetic force on a moving charged particle.
The formula is given by:
|F| = q * |v| * |B| * sinθ
where:
|F| is the magnitude of the force,
q is the charge of the particle (in this case, the charge of the proton is 1.6 × 10^-19 C),
|v| is the magnitude of the velocity of the proton (6.8 × 10^6 m/s),
|B| is the magnitude of the magnetic field, and
θ is the angle between the velocity vector and the magnetic field vector.
From the given information, we know that |F| = 8.6 × 10^-14 N, |v| = 6.8 × 10^6 m/s, and the direction of the magnetic force is due south. Since the force is acting vertically downward, we can infer that the angle between the velocity vector and the magnetic field vector is 90 degrees (π/2 radians).
By rearranging the formula, we can solve for |B|:
|B| = |F| / (q * |v| * sinθ)
Substituting the known values:
|B| = (8.6 × 10^-14 N) / ((1.6 × 10^-19 C) * (6.8 × 10^6 m/s) * sin(π/2))
Calculating this expression:
|B| = 6.25 × 10^-5 T
Since the force is acting downward, the magnetic field is in the opposite direction, which is upward. Therefore, the direction of the magnetic field causing the force is up.