Question

A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular instant, it is moving with a speed of 5.00 ✕ 106 m/s through a magnetic field. At this instant, its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location. The magnitude of the field is 0.220 T. What is the magnitude of the magnetic force (in N) on the ion? explain step by step

Answer 1

`F_m =1.4* 10^(-12)\ N`

Step-by-step explanation:

Given:

charge on a beryllium-9 ion,
`Q=3.2* 10^(-19)\ C`

mass of Be-9 ion,
`m=1.5* 10^(-26)\kg`

instantaneous speed of the ion,
`v=5* 10^6\ m.s^(-1)`

angle between the ion-velocity and the magnetic field lines,
`\theta=61^(\circ)`

magnitude of magnetic field intensity,
`B=0.22\ T`

Now as we know that the magnetic force on a charge is given as:

`F_m=Q.v* B`

`F_m=Q.v.B\sin theta`

`F_m=3.2* 10^(-19)* 5* 10^6* 0.22* \sin 61^(\circ)`

`F_m =1.4* 10^(-12)\ N`

Answer 2

Magnetic force,
`F = 3.52* 10^(-13)\ N`

Step-by-step explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton,
`q=2* 1.6* 10^(-19)\ C=3.2* 10^(-19)\ C`

Speed of the ion in the magnetic field,
`v=5* 10^6\ m/s`

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

`F=qvB\\\\F=3.2* 10^(-19)* 5* 10^6* 0.22\\\\F=3.52* 10^(-13)\ N`

So, the magnitude of magnetic force on the ion is
`3.52* 10^(-13)\ N`.