Question

A charged particle is accelerated from rest through a potential difference of magnitude |ΔV|. After exiting the potential difference at an emission point, the particle enters a region of uniform magnetic field. The magnetic field is perpendicular to the particle's velocity, and the particle travels along a complete circular path. The particle's mass is 2.10 ✕ 10−16 kg, its charge is 26.0 nC, and the magnetic field magnitude is 0.600 T. The particle's circular path, as it returns to the emission point, encloses a magnetic flux of 15.0 µWb.

a. What is the speed in (m/s) of the particle when it is in the region of the magnetic field?
b. What is the magnitude of the potential difference through which the particle was accelerated?

Answer 1

a

The speed of the particle is
`v = 209485.71 m/s`

b

The potential difference is
`\Delta V = 177.2 \ V`

Step-by-step explanation:

From the question we are told that

The mass of the particle is
`m = 2.10 *10^(-16) kg`

The charge on the particle is
`q = 26.0 nC = 26.0 *10^(-9) C`

The magnitude of the magnetic field is
`B = 0.600 T`

The magnetic flux is
`\O = 15.0 \mu Wb = 15.0 *10^(-6) Wb`

The magnetic flux is mathematically represented as

`\O = B *A`

Where A is the the area mathematically represented as

`A = \pi r^2`

Substituting this into the equation w have

`\O = B (\pi r^2 )`

Making r the subject of the formula

`r = \sqrt{(\O)/(B \pi) }`

Substituting value

`r = \sqrt{(15 *10^(-6))/(3.142 * 0.6) }`

`r = 2.82 *10^(-3)m`

For the particle to form a circular path the magnetic force the partial experience inside the magnetic must be equal to the centripetal force of the particle and this is mathematically represented as

`F_q = F_c`

Where
`F_q = q B v`

and
`F_c = (mv^2 )/(r)`

Substituting this into the equation above

`qBv = (mv^2)/(r)`

making v the subject

`v = (r q B)/(m)`

substituting values

`v = (2.82 * 10^(-3) * 26 *10^(-9) 0.6)/(2.10*10^(-16))`

`v = 209485.71 m/s`

The potential energy of the particle before entering the magnetic field is equal to the kinetic energy in the magnetic field

This is mathematically represented as

`PE = KE`

Where
`PE = q * \Delta V`

and
`KE = (1)/(2) mv^2`

Substituting into the equation above

`q \Delta V = (1)/(2) mv^2`

Making the potential difference the subject

`\Delta V = (mv^2)/(2 q)`

`\Delta V = \frac{2.16 * 10^ {-16} * (209485.71)^2 }{2 * 26 *10^(-9)}`

`\Delta V = 177.2 \ V`