Chapter 26, Problem 41P

(a)

To determine

The time required for $Ni$ ion to complete the semicircular path.

Answer to Problem 41P

  $15.7\times {10}^{-6}\text{s}$

Explanation of Solution

Given:

Magnitude of charge on an ion, $q=1.602\times {10}^{-19}\text{C}$

Magnitude of magnetic field, $B=0.120T$

Magnitude of velocity of ion $=v$

Mass of the ion $Ni$$=m_{58}=(58)(1.6606\times {10}^{-27})\text{ kg }$

Angle between the velocity of ion and magnetic field, $\theta =90\text{ deg}$

Radius of the orbit of ion $Ni$$=r$

Time taken to complete the semicircle by ion $Ni$$=t_{58}$

Distance traveled to complete the semicircle $=d$

Formula Used:

Time taken is given as

  $\begin{array}{l}\text{Time}=\frac{\text{Distance}}{\text{speed}}\\ t=\frac{d}{v}\end{array}$

Magnetic force on the ion moving in magnetic field is given

  $F=qvB\mathrm{Sin}\theta$

Centripetal force is given as

  $F_c=\frac{m{v}^2}{r}$

Calculation:

Magnetic force on the ion moving in magnetic field is given

  $F=qvB\mathrm{Sin}\theta$

For the ion to move in a circular orbit, the necessary centripetal force is provided by the magnetic force acting on it. Hence

  $\begin{array}{l}F=F_c\\ qvB\mathrm{Sin}\theta =\frac{m{v}^2}{r}\\ r=\frac{mv\mathrm{Sin}\theta }{qB}\\ \text{since }\theta =90,\\ r=\frac{mv\mathrm{Sin}90}{qB}\\ r=\frac{mv}{qB}\text{ Eq-1}\end{array}$

Distance traveled to complete the semicircle is same as the circumference of semicircle

  $\begin{array}{l}\text{distance traveled = Circumference of semicircle}\\ d=\pi r\end{array}$

Time taken to complete the semicircle is given as

  $\begin{array}{l}t=\frac{d}{v}\\ t=\frac{\pi r}{v}\\ \text{using Eq-1}\\ t=\frac{\pi }{v}\left(\frac{mv}{qB}\right)\\ t=\frac{\pi m}{qB}\end{array}$

So time taken by ion $Ni$ to complete the semicircle is given as

  $\begin{array}{l}{t}_{58}=\frac{\pi m_{58}}{qB}\\ t_{58}=\frac{\pi (58)(1.6606\times {10}^{-27})\text{ kg }}{(1.6\times {10}^{-19})(0.120)}\\ t_{58}=15.7\times {10}^{-6}\text{s}\end{array}$

Conclusion:

The time taken by ion $Ni$ comes out to be $15.7\times {10}^{-6}\text{s}$

(b)

To determine

The time required for $Ni$ ion to complete the semicircular path.

Answer to Problem 41P

  $16.4\times {10}^{-6}s$

Explanation of Solution

Given:

Magnitude of charge on an ion, $q=1.602\times {10}^{-19}\text{C}$

Magnitude of magnetic field, $B=0.120T$

Magnitude of velocity of ion $=v$

Mass of the ion $Ni$

  $m_{60}=(60)(1.6606\times {10}^{-27})\text{ kg }$

Angle between the velocity of ion and magnetic field $=\theta =90\text{ deg}$

Radius of the orbit of ion $Ni$$=r$

Time taken to complete the semicircle by ion $Ni$$=t_{60}$

Distance traveled to complete the semicircle $=d$

Formula Used:

Time taken is given as

  $\begin{array}{l}\text{Time}=\frac{\text{Distance}}{\text{speed}}\\ t=\frac{d}{v}\end{array}$

Magnetic force on the ion moving in magnetic field is given

  $F=qvB\mathrm{Sin}\theta$

Centripetal force is given as

  $F_c=\frac{m{v}^2}{r}$

Calculation:

Magnetic force on the ion moving in magnetic field is given

  $F=qvB\mathrm{Sin}\theta$

For the ion to move in a circular orbit, the necessary centripetal force is provided by the magnetic force acting on it. Hence

  $\begin{array}{l}F=F_c\\ qvB\mathrm{Sin}\theta =\frac{m{v}^2}{r}\\ r=\frac{mv\mathrm{Sin}\theta }{qB}\\ \text{since }\theta =90,\\ r=\frac{mv\mathrm{Sin}90}{qB}\\ r=\frac{mv}{qB}\text{ Eq-1}\end{array}$

Distance traveled to complete the semicircle is same as the circumference of semicircle

  $\begin{array}{l}\text{distance traveled = Circumference of semicircle}\\ d=\pi r\end{array}$

Time taken to complete the semicircle is given as

  $\begin{array}{l}t=\frac{d}{v}\\ t=\frac{\pi r}{v}\\ \text{using Eq-1}\\ t=\frac{\pi }{v}\left(\frac{mv}{qB}\right)\\ t=\frac{\pi m}{qB}\end{array}$

So time taken by ion $Ni$ to complete the semicircle is given as

  $\begin{array}{l}{t}_{60}=\frac{\pi m_{60}}{qB}\\ t_{60}=\frac{\pi (60)(1.6606\times {10}^{-27})\text{ kg }}{(1.6\times {10}^{-19})(0.120)}\\ t_{60}=16.3\times {10}^{-6}\text{s}\end{array}$

Conclusion:

The time taken by ion $Ni$ comes out to be $16.3\times {10}^{-6}\text{s}$ .