Question

Figure 28-29 shows 11 paths through a region of uniform magnetic field. One path is a straight line; the rest are half-circles. Table 28-4 gives the masses, charges, and speeds of 11 particles that take these paths through the field in the directions shown. Which path in the figure corresponds to which particle in the table? (The direction of the magnetic field can be determined by means of one of the paths, which is unique.)

Short answer

Particles 1 to 11 corresponds to path i, e, c, a, g, j, d, b, h, f, k respectively.

Step-by-step solution

Given

Corresponding mass, velocity and charge of different particles are given in table.

Determining the concept

Using the relation28-16for radius find the radius of different particles and analyzing the given diagram find the corresponding paths of different particles.

Formulae are as follow:

Where, r is radius, B is magnetic field, v is velocity, m is mass,q is charge on particle.

Determining the paths in the figure corresponds to which particle in table

$m=m_l=8N_1=-q$

Using equation 28-16, find the radius of different paths.

For particle 1

m=2m,V=v,q=q

The path i shows the radius r=2. Hence particle 1 shows the path i

For particle 2,

The path e shows the radius r=0.5.

Hence, particle 2 shows the path e.

For particle 3,

The path c shows the radius r=1

Hence, particle 3 shows the path c.

For particle 4,,

m=3m,V=3v,q=3q

The path a shows the radius r=3

Hence, particle 4 shows the path a.

For particle 5

The path g shows the radius r=4

Hence, particle 5 shows the path g.

For particle 6,

The path j shows the radius r=-2

Hence, particle 6 shows the path j.

For particle 7,

The path d shows the radius r=0.25

Hence, particle 7 shows the path d

For particle 8

The path b shows the radius r=-1

Hence, particle 8 shows the path b

For particle 9,

The path h shows the radius r=-3

Hence, particle 9 shows the path h

For particle 10

$m=m_l=8N_1=-q$
$r=\frac{m\left(8v\right)}{-2qB}=-4\left(\frac{m}{qB}\right)$

The path shows the radius r=-4

Hence, particle 10 shows the path f

For particle 11

$m=3m_{|}=8_{1}q=$0

$r=\frac{\left(3m\right)\left(3w\right)}{\left(0\right)B}=0$

The path shows the radius r=0

Hence, particle 11 shows the path k.

Hence, particles 1 to 11 corresponds to path i, e, c, a, g, j, d, b, h, f, k respectively.