A particle is shot into a magnetic field region as shown in the figure.

Question

A particle is shot into a magnetic field region as shown in the figure.

If the particle is shot in with a speed of
𝑣=392 m/s, has a charge of
𝑞=33.6 𝜇C and a mass of 𝑚=0.3 mg, how strong must the magnetic field be so the particle just misses the right plate, given that the plates are separated by a distance b
𝑑=0.16 m ?

Answer 1

To determine the required strength of the magnetic field so that the particle just misses the right plate, we need to consider the forces acting on the particle.

The force experienced by a charged particle moving in a magnetic field is given by the formula:

𝐹 = 𝑞(𝑣 × 𝐵)

where 𝑞 is the charge of the particle, 𝑣 is the velocity of the particle, 𝐵 is the magnetic field, and × represents the cross product.

For the particle to miss the right plate, the magnetic force must provide the necessary centripetal force to keep the particle moving in a circular path. The centripetal force is given by:

𝐹𝑐 = 𝑚(𝑣²/𝑟)

where 𝑚 is the mass of the particle, 𝑣 is the velocity, and 𝑟 is the radius of the circular path.

In this case, the radius of the circular path is equal to the distance between the plates, 𝑑.

Setting the magnetic force equal to the centripetal force, we have:

𝑞(𝑣 × 𝐵) = 𝑚(𝑣²/𝑟)

Replacing 𝑟 with 𝑑, we have:

𝑞(𝑣 × 𝐵) = 𝑚(𝑣²/𝑑)

Rearranging the equation to solve for 𝐵, we get:

𝐵 = (𝑚(𝑣²/𝑑))/𝑞𝑣

Substituting the given values into the equation and solving for 𝐵:

𝑚 = 0.3 mg = 0.3 x 10^-6 kg (converting from mg to kg)
𝑣 = 392 m/s
𝑞 = 33.6 μC = 33.6 x 10^-6 C (converting from μC to C)
𝑑 = 0.16 m

Plugging in the values:

𝐵 = (0.3 x 10^-6 kg)((392 m/s)²/(0.16 m))/(33.6 x 10^-6 C)(392 m/s)

Simplifying:

𝐵 = (0.3 x 10^-6 kg)(153,664 m²/s²)/(13.3632 x 10^-6 C·m/s)

Calculating:

𝐵 = 14.2 T

Therefore, to just miss the right plate, the magnetic field needs to be 14.2 Tesla (T).