Question

This problem has been solved! See the answer A proton with a speed of 3.5x10^6 m/s is shot into a region between two plates that are separated by distance of 0.23 m. A magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field so the proton just misses colliding with the opposite plate?

Answer 1

Answer:

The magnitude of the magnetic field 'B' is 0.16 Tesla.

Explanation:

The magnitude of the magnetic field 'B' can be determined by;

             B = $\frac{mV}{qR}$

where: m is the mass of proton, V is its speed , q is the charge of proton and R is the distance between the plates.

Given that: speed 'V' of the proton = 3.5 × $10^{6} ms^{-2}$, distance 'R' between the plates = 0.23m, the charge 'q' on proton = 1.9 × $10^{-19}$ C and mass of proton = 1.67 × $10^{-27}$Kg.

Thus,

  B = (1.67 × $10^{-27}$ × 3.5 ×$10^{6}$) ÷ (1.6 × $10^{-19}$ ×  0.23)

      = $\frac{5.845 * 10^{-21} }{3.68 * 10^{-20} }$

      = 0.15883

 B = 0.16 Tesla

The magnitude of the magnetic field 'B' is 0.16 Tesla.