Question

After being accelerated to a speed of1.13×10⁵ m/s, the particle enters a uniform magnetic field of strength 0.900 T and travels in a circle of radius 34.0 cm (determined by observing where it hits the screen as shown in the figure). The results of this experiment allow one to find m/q.

Find the ratio m/q for this particle.

Answer 1

Concepts and reason

The concepts used in this problem are magnetic force and the centripetal force. The ratio of mass to the charge can be calculated using the expressions of magnetic force and the centripetal force.

Fundamentals

Centripetal Force:

The linear acceleration is the rate of change of linear velocity but the rate of change of tangential velocity is the centripetal acceleration. The centripetal acceleration makes an object to move in a curved path. The force acts in this motion is centripetal force

The expression for centripetal force is:

${F_c} = \frac{{m{v^2}}}{r}$

Here, $m$ is the mass, ${a_c}$ is the centripetal acceleration, $v$ is the velocity and $r$ is the radial distance.

Magnetic force:

The magnetic force is perpendicular to the magnetic field and the velocity of the charge. The magnitude of magnetic force is given by:

${F_m} = qvB\sin \theta$

Here, $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field and $\theta$ is the angle between velocity and the magnetic field.

The forces acting on the system are magnetic force and centripetal force, both the force should be balanced.

$\begin{array}{c}\\{F_m} = {F_c}\\\\qvB\sin \theta = \frac{{m{v^2}}}{r}\\\\\frac{m}{q} = \frac{{Br\sin \theta }}{v}\\\end{array}$

Here, the velocity vector is perpendicular to the magnetic field vector. So, angle is ${90^{\rm{o}}}$ .

Substitute ${90^{\rm{o}}}$ for $\theta$ .

$\begin{array}{c}\\\frac{m}{q} = \frac{{Br\sin {{90}^{\rm{o}}}}}{v}\\\\\frac{m}{q} = \frac{{Br}}{v}\\\end{array}$

The ratio of mass to the charge is:

$\frac{m}{q} = \frac{{Br}}{v}$

Substitute $0.900{\rm{ T}}$ for $B$ , $34{\rm{ cm}}$ for $r$ and $1.13 \times {10^5}{\rm{ m/s}}$ for $v$ .

$\begin{array}{c}\\\frac{m}{q} = \frac{{\left( {0.900{\rm{ T}}} \right)\left( {34{\rm{ cm}}} \right)}}{{1.13 \times {{10}^5}{\rm{ m/s}}}}\\\\ = \frac{{\left( {0.900{\rm{ T}}} \right)\left( {34{\rm{ cm}} \times \frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)}}{{1.13 \times {{10}^5}{\rm{ m/s}}}}\\\\ = \frac{{\left( {0.900{\rm{ T}}} \right)\left( {34 \times {{10}^{ - 2}}{\rm{ m}}} \right)}}{{1.13 \times {{10}^5}{\rm{ m/s}}}}\\\end{array}$

The equation is further simplified as:

$\begin{array}{c}\\\frac{m}{q} = \frac{{\left( {0.900{\rm{ T}}} \right)\left( {34 \times {{10}^{ - 2}}{\rm{ m}}} \right)}}{{1.13 \times {{10}^5}{\rm{ m/s}}}}\\\\ = \frac{{0.306{\rm{ T}} \cdot {\rm{m}}}}{{1.13 \times {{10}^5}{\rm{ m/s}}}}\\\\ = 2.707 \times {10^{ - 6}}{\rm{ kg/C}}\\\\ = {\bf{2}}{\bf{.71 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ kg/C}}\\\end{array}$

Ans:

The ratio of mass to the charge of a particle is ${\bf{2}}{\bf{.71 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ kg/C}}$ .