Question

An electron at point A in the figure has a speed of 1.42×10∧6 m/s.

(a) Find the direction of the magnetic field that will cause the electron to follow the semicircular path from A to B.

(b) Find the magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B.

(c) Find the time required for the electron to move from A to B.

(d) Find the direction of the magnetic field, that would be needed if the particle were a proton instead of an electron?

(e) Find the magnitude of the magnetic field, that would be needed if the particle were a proton instead of an electron?

Answer 1

Concepts and reason

The concepts required to solve this problem are the right-hand rule, the magnetic force on a moving charged particle, and centripetal force. Determine the direction of the magnetic field for both electron and proton with the help of the right-hand rule. Calculate the magnetic field of electrons and protons by equating the centripetal force and magnetic force. Calculate the time required for electrons to move along the semicircular arc by using the expression of the time period.

Fundamentals

Right-hand Rule: Place your middle figure, index figure, and thumb of your right hand in such a way that they are perpendicular to each other. Now, if the index finger points in the direction of the velocity vector and the middle finger points in the direction of the magnetic field vector, then the thumb will point in the direction of the force vector. The magnitude of magnetic force \(F_{\mathrm{B}}\) on a charged particle of mass \(\mathrm{m}\) moving at a speed \(v\) in a magnetic field \(\mathrm{B}\) perpendicular to the direction of velocity is given as follows:

\(F_{\mathrm{B}}=q v B\)

The centripetal force \(F_{C}\) on a particle of mass \(\mathrm{m}\) moving at a speed \(v\) in a circle of radius \(r\) is given as follows:

\(F_{\mathrm{C}}=\frac{m v^{2}}{r}\) Time-period of revolution in a circular path is given by the following expression:

\(T=\frac{2 \pi r}{v}\)

Here, \(T\) is the time period of revolution.

(a) The trajectory of the electron will be semicircular only if the magnetic field will be perpendicular to the direction of velocity. Place your index fingers in the direction of velocity and thumb towards the center of the semicircle then the middle figure will point in the direction of the magnetic field. The only direction of the magnetic field which will let the thumb point towards the center according, to the formula, \(\vec{F}=q(\vec{v} \times \vec{B})\) is that directed out of the paper. But, since the particle is an electron the direction of the magnetic field will be directed into the page of the paper.

The direction of the force in \(\vec{F}=q(\vec{v} \times \vec{B})\) finds out with the help of the right-hand rule. It shows the direction of the force on a positive charge. If the charge will be negative then the direction of the force will change. That is, opposite to the direction of the force on a positive charge.

(b) The magnitude of magnetic force \(F_{\mathrm{B}}\) on the electron is given as follows:

\(F_{\mathrm{B}}=q v B\)

The centripetal force \(F_{\mathrm{C}}\) on the electron is given as follows:

\(F_{\mathrm{C}}=\frac{m v^{2}}{r}\)

Here, \(m\) is the mass, \(v\) is the velocity and \(r\) is the radius of the semicircular path. Equating centripetal force with a magnitude of magnetic force and solve for \(\mathrm{B}\). \(F_{\mathrm{B}}=F_{\mathrm{C}}\)

Substitute \(q v B\) for \(F_{\mathrm{B}}\) and \(\frac{m v^{2}}{r}\) for \(F_{\mathrm{C}}\)

\(q v B=\frac{m v^{2}}{r}\)

\(\boldsymbol{B}=\frac{m v}{q r}\)

The radius r of the semicircular arc is half of the diameter \(\mathrm{d}\) of the arc. \(r=\frac{d}{2}\)

Substitute \(10.0 \mathrm{~cm}\) for \(d\) in the above equation.

\(r=\frac{10.0 \mathrm{~cm}}{2}\)

\(=(5.00 \mathrm{~cm})\left(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\right)\)

\(=0.05 \mathrm{~m}\)

Substitute \(9.1 \times 10^{-31} \mathrm{~kg}\) for \(m, 1.42 \times 10^{6} \mathrm{~m} / \mathrm{s}\) for \(v, 1.6 \times 10^{-19} \mathrm{C}\) for \(q\) and \(0.05 \mathrm{~m}\) for \(r\) in \(B=\frac{m v}{q r}\).

\(B=\frac{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(1.42 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)}{\left(1.6 \times 10^{-19} \mathrm{C}\right)(0.05 \mathrm{~m})}\)

\(=\left(1.62 \times 10^{-4} \mathrm{~T}\right)\left(\frac{10^{6} \mu \mathrm{T}}{1 \mathrm{~T}}\right)\)

$$ =162 \mu \mathrm{T} $$

Since the electron follows a circular path, the magnitude of the magnetic force on the electron must be equal to centripetal force. \(F_{B}=F_{C}\)

The diameter of the semicircular path of the electron is \(10.0 \mathrm{~cm}\) and formula \(B=\frac{m v}{q r}\) requires the radius. Hence, the radius is calculated using \(r=\frac{d}{2}\).

(c) The expression for time-period is as follows:

\(T=\frac{2 \pi r}{v}\)

It is the expression for time-period for one complete revolution. Since the electron has completed only half of a circle hence, the time required to complete it will be half of \(\mathrm{T}\). \(t=\frac{T}{2}\)

Substitute \(\frac{2 \pi r}{v}\) for \(\mathrm{T}\). \(t=\frac{\pi r}{v}\)

Substitute \(0.05 \mathrm{~m}\) for \(r\) and \(1.42 \times 10^{6} \mathrm{~m} / \mathrm{s}\) for \(v\) in the above equation \(t=\frac{\pi r}{v}\) and solve for \(\mathrm{t}\).

$$ \begin{array}{c} t=\frac{\pi(0.05 \mathrm{~m})}{1.42 \times 10^{6} \mathrm{~m} / \mathrm{s}} \\ =\left(0.111 \times 10^{-6} \mathrm{~s}\right)\left(\frac{10^{6} \mu \mathrm{s}}{1 \mathrm{~s}}\right) \\ =0.111 \mu \mathrm{s} \end{array} $$

The time period of one complete circular revolution is \(\frac{2 \pi r}{v}\). That is \(T\) is the time required to complete \(2 \pi r\) distance. Since from \(\mathrm{A}\) to \(\mathrm{B}\) is a semicircular arc, the distance will be \(\pi r\). Hence, to cover \(\pi r\) distance, the required time will be \(t=\frac{\pi r}{v}\).

(d) If the particle would be a proton the direction of the magnetic field will be opposite to the direction that of the electron. The direction of magnetic field for an electron was on the page of paper. Hence, for protons, the direction of the magnetic field will be directed out of the page of the paper.

According to the right-hand rule, the direction of the magnetic field on proton would be pointed out of the page. The formula \(\vec{F}=q(\vec{v} \times \vec{B})\) indicates the direction of force on a positive charge. If \(q\) is negative then force will reverse the direction. If you require to keep the same direction of force then you must change the direction of the magnetic field. The following expression will explain this fact. \(\vec{F}=(-q)(\vec{v} \times(-\vec{B}))\)

Both the negative signs will cancel each other to keep the direction of force same.

(e) The expression for magnetic field B by equating magnetic force and centripetal force is given as follows:

\(B=\frac{m v}{q r}\)

Substitute \(1.67 \times 10^{-27} \mathrm{~kg}\) for \(m, 1.42 \times 10^{6} \mathrm{~m} / \mathrm{s}\) for \(v, 1.6 \times 10^{-19} \mathrm{C}\) for \(q\), and \(0.05 \mathrm{~m}\) for \(r\) in the above equation.

$$ \begin{array}{c} B=\frac{\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\left(1.42 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)}{\left(1.6 \times 10^{-19} \mathrm{C}\right)(0.05 \mathrm{~m})} \\ =0.296 \mathrm{~T} \end{array} $$

The magnetic force \(F_{C}\) on proton is equal to the centripetal force \(F_{B}\) on proton because it follows a circular motion. \(F_{B}=F_{C}\)

The magnitude of the magnetic field required for protons is greater than the magnitude of the magnetic field required for electrons. It is due to the heavier mass of protons than electrons.

Part a The direction of the magnetic field on the electron is into the page.

Part \(\mathrm{b}\) The magnitude of the magnetic field on the electron is \(162 \mu \mathrm{T}\).

Part c The time required by the electron to travel from \(\mathrm{A}\) to \(\mathrm{B}\) is \(0.111 \mu \mathrm{s}\).

Part d The direction of the magnetic field on the proton must be out of the page.

Part e The magnitude of the magnetic field for proton is \(0.296 \mathrm{~T}\).