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

Answer:

$F_m =1.4\times 10^{-12}\ N$

Explanation:

Given:

charge on a beryllium-9 ion, $Q=3.2\times 10^{-19}\ C$

mass of Be-9 ion, $m=1.5\times 10^{-26}\kg$

instantaneous speed of the ion, $v=5\times 10^6\ m.s^{-1}$

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:

$F_m=Q.v\times B$

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

$F_m=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\times \sin 61^{\circ}$

$F_m =1.4\times 10^{-12}\ N$

Answer 2

Answer:

Magnetic force, $F = 3.52\times 10^{-13}\ N$

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, $q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C$

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

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 :

$F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N$

So, the magnitude of magnetic force on the ion is $3.52\times 10^{-13}\ N$.