IB Physics HL D.3.2 Charge in magnetic fields

[Maximum number: 5]

A rod, R , lies perpendicular to a uniform magnetic field B of strength 0.50 T directed into the plane of the page. R is connected to a circuit and the electric current $I_{\mathrm{R}}$ is 2.0 A .

(a)

An electron is moving at constant velocity $v=9.0 \times 10^{5} \mathrm{~ms}^{-1}$ parallel to a rod T , at a perpendicular distance d. T is connected to a cell via a switch. When the switch is closed the electron's path curves with initial radius r. The magnetic field strength at the position of the electron at d is $4.0 \times 10^{-6} \mathrm{~T}$ .

[ 5 ]

(i)

Calculate the initial radius of the electron's path.

[ 3 ]

use of $\mathrm{qvB}=\mathrm{mv}^{2} / \mathrm{r}$

OR
$\begin{aligned} & r=\frac{m v}{q B} & 1.6 \times 10^{-19} \times 4.0 \times 10^{-6}=\frac{9.11 \times 10^{-31} \times 9 \times 10^{5}}{r} \end{aligned}$

OR
$\begin{aligned} & r=\frac{\left(9.11 x 10^{-31}\right)\left(9 x 10^{5}\right)}{\left[\left(1.6 \times 10^{-19}\right)\left(4 x 10^{-6}\right)\right]} & r=1.28 \text { OR } 1.3 \text { «m» } \end{aligned}$

(ii)

Explain the subsequent change in the radius of the electron's path.

[ 2 ]

IB Physics HL D 3 2 Charge in Magnetic Fields Questions

Exam points

IB Physics HL D 3 2 Charge in Magnetic Fields Questions question 1

Question (a)

Question (i)

Question (ii)