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Q11Q

Expert-verifiedFound in: Page 828

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

** In Fig. 28-30, a charged particle enters a uniform magnetic field with speed ${\mathbf{v}}_{0}$** **, moves through a halfcirclein time${\mathbf{T}}_{0}$ ** **, and then leaves the field**

**. (a) Is the charge positive or negative? **

**(b) Is the final speed of the particle greater than, less than, or equal to ${{\mathbf{v}}}_{{\text{0}}}$** **? **

**(c) If the initial speed had been${\text{0.5}}{{\mathbf{v}}}_{{\text{0}}}$ ** **, would the time spent in field have been greater than, less than, or equal to ${{\mathbf{T}}}_{{\text{0}}}$** **? **

**(d) Would the path have been ahalf-circle, more than a half-circle, or less than a half-circle?**

- The charge on the particle is negative.
- The final speed of the particle is equal to ${v}_{0}.$
- If the initial speed is $0.5{v}_{0}$, then the time spent in the magnetic field $\overrightarrow{B}$ is equal to ${T}_{0}.$
- The path would have been a half circle.

Consider the formula for the radius of the charge as:

${\mathbf{r}}{\mathbf{=}}\frac{\mathrm{mv}}{\mathbf{\left|}\mathbf{q}\mathbf{\right|}\mathbf{B}}$

Here, *r* is radius, *B* is magnetic field, *v* is velocity, *m* is mass,*q* is charge on particle

The charge on the particle:

In order to get the force on the particle downwardaccording to the right hand rule, the charge on the particle must be negative.

Hence, the charge on the particle is negative.

The final speed of the particle greater than, less than, or equal to ${v}_{0}$ :

Magnetic field does not do any work on the charge particle. It only changes the direction of velocity in magnetic field region.

Hence, the speed of the particle does not change.

Hence, its final speed is the same as the initial speed.

If the initial speed is $0.5{v}_{0}$ , then whether the time spent in the magnetic field $\overrightarrow{B}$ is greater than, less than, or equal to ${T}_{0}$:

$r=\frac{mv}{\left|q\right|B}$

If velocity is $\frac{1}{2}{v}_{0},$ then using the above relation,

$r=\frac{1}{2}\left(\frac{mv}{qB}\right)$

Hence, radius will get halved.

Consider the formula:

$T=\frac{2\pi r}{v}$

Since the speed is halved and the radius is halved, the period will not change. It will be the same as the initial time ${T}_{0}.$

Hence, if the initial speed is $0.5{v}_{0}$ , then the time spent in the magnetic field $\overrightarrow{B}$ is equal to ${T}_{0}.$

Path is half circle, more than half circle, or less than a half circle:

The path will be half circle. Speed does not change the direction; only the magnetic force changes the direction.

Hence, the path would have been a half circle.

Therefore, use the right hand rule to find the force on the particle and usethe relation of radius and magnetic field to find the radius of different particles.

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