A proton traveling at 25.6° with respect to the direction of a magnetic field of strength 3.35 mT experiences a magnetic force of 8.02 × 1017 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts. (a) Number Units (b) Number Units
Q: A proton traveling at 26.6° with respect to the direction of a magnetic field of strength 3.10 mT…
A:
Q: A proton traveling at 20.4° with respect to the direction of a magnetic field of strength 2.83 mT…
A:
Q: An ion with a charge of 1.6 x 10^-19 C moves at speed 1.0 x 10^6 m/s into and perpendicular to the…
A:
Q: A proton traveling at 21.8° with respect to the direction of a magnetic field of strength 3.43 mT…
A:
Q: An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.29 T. If the…
A: Given: The magnitude of the magnetic field, B = 0.29 T The kinetic energy of the electron, K.E. =…
Q: An electron moves with a speed of 5.0 × 104 m/s perpendicular to a uniform magnetic field of…
A: Given: Speed of the electron =5.0×104 m/s Magnitude of the uniform magnetic field =0.20 T Charge of…
Q: A proton traveling at 24.6° with respect to the direction of a magnetic field of strength 2.50 mT…
A: Magnetic field B = 2.50 mT = 2.50 × 10- 3 T Magnetic force F = 6.36 ×10- 17 N Velocity angle with…
Q: A proton traveling at 32.5° with respect to the direction of a magnetic field of strength 3.51 mT…
A: Given B= 3.51 mT F= 5.71 x 10-17 N angle between the direction of V and B = 32.50 mass of…
Q: An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.265 T. If the…
A: Given: B= 0.265 T E= 3.80×10-19 J R=? m= 9.1×10-31 kg a) E= 12mv2 Rearrange it for V V=…
Q: An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.29 T. If the…
A: a. the kinetic energy of the electron is
Q: A proton traveling at 17.9° with respect to the direction of a magnetic field of strength 3.56 mT…
A: The magnetic field strength: The angle between the direction of the magnetic field and the direction…
Q: A proton traveling at 20.0° with respect to the direction of a magnetic field of strength 3.29 mT…
A: It is given that a proton traveling at 20.0° with respect to the direction of a magnetic field of…
Q: A proton traveling at 27.4° with respect to the direction of a magnetic field of strength 3.78 mT…
A:
Q: A proton traveling at 24.0° with respect to the direction of a magnetic field of strength 2.72 mT…
A: Angle made by the proton with respect to the direction of magnetic field (θ) = 24omagnetic field…
Q: A proton traveling at 27.8° with respect to the direction of a magnetic field of strength 3.25 mT…
A: Magnetic field B = 3.25 mT = 3.25 × 10- 3 T Magnetic force F = 8.93 × 10- 17 N Angle between…
Q: A proton traveling at 31.5° with respect to the direction of a magnetic field of strength 3.43 mT…
A: Given,B=3.43 mTF=9.69×10-17Nθ=31.5°
Q: A proton traveling at 28.7° with respect to the direction of a magnetic field of strength 3.23 mT…
A:
Q: A proton traveling at 18.6° with respect to the direction of a magnetic field of strength 2.88 mT…
A: The angle at which the proton travels is given as, θ=18.6o. The strength of the magnetic field is…
Q: A proton traveling at 28.3° with respect to the direction of a magnetic field of strength 2.48 mT…
A: We use F = qvBsinθ to get, v=FqBsinθ and, E = 12mv2
Q: A drill bit is able to reach 2000 rpm in 0.50 s. Assuming a constant angular acceleration, how many…
A: Given: The final angular speed of the drill is dω=2000 rpm= 2000×2π60 s=209.44 rads. Time required…
Q: A proton traveling at 21.2° with respect to the direction of a magnetic field of strength 3.61 mT…
A:
Q: A proton traveling at 18.6° with respect to the direction of a magnetic field of strength 2.84 mT…
A:
Q: An electron moves across Earth’s equator at a speed of 2.5 x 106 m/s and a direction 35o north of…
A: Given that, Earth's magnetic field B=0.1×10-4 TCharge of electron q=1.6×10-19 CSpeed of electron…
Q: A proton traveling at 22.9° with respect to the direction of a magnetic field of strength 2.41 mT…
A:
Q: A proton traveling at 31.9° with respect to the direction of a magnetic field of strength 3.30 mT…
A:
Q: A proton traveling at 20.8° with respect to the direction of a magnetic field of strength 3.12 mT…
A:
Q: A proton traveling at 21.1° with respect to the direction of a magnetic field of strength 3.78 mT…
A:
Q: A proton traveling at 26.6° with respect to the direction of a magnetic field of strength 3.62 mT…
A:
Q: A proton traveling at 20.0° with respect to the direction of a magnetic field of strength 3.29 mT…
A:
Q: A proton moving perpendicular to a magnetic field of strength 3.2 mT experiences a force due to the…
A: proton is moving perpendicular which means angle θ =900 Force F =6.0 × 10-21 N Magnetic field B =3.2…
Q: A proton traveling at 32.9° with respect to the direction of a magnetic field of strength 2.27 mT…
A:
Q: roton traveling at 31.6° with respect to the direction of a magnetic field of strength 3.49 mT…
A:
Q: A proton traveling at 17.3° with respect to the direction of a magnetic field of strength 3.49 mT…
A:
Q: A proton traveling at 33.3° with respect to the direction of a magnetic field of strength 3.14 mT…
A: θ = 33.3oB = 3.14 mT F = 6.73×10-17 N
Q: A proton traveling at 19.2° with respect to the direction of a magnetic field of strength 3.00 mT…
A: Given data: Angle between velocity and magnetic field (θ) = 19.2° Magnetic field (B) = 3.00 mT =…
Q: A proton traveling at 27.4° with respect to the direction of a magnetic field of strength 3.78 mT…
A:
Q: A proton traveling at 25.5° with respect to the direction of a magnetic field of strength 3.48 mT…
A:
Q: A proton traveling at 25.6° with respect to the direction of a magnetic field of strength 3.35 mT…
A:
Q: A proton traveling at 24.2° with respect to the direction of a magnetic field of strength 2.22 mT…
A:
Q: A proton traveling at 23.9 degrees with respect to the direction of a magnetic field of strength…
A: a. the force due to magnetic field can be expressed as
Step by step
Solved in 2 steps with 2 images