Question

What is the initial direction of deflection for the charged particles entering the magnetic fields shown in the figure? (Conceptual Question)

Answer 1

A particle moves in a region of uniform magnetic field with magnitude (B)(B)isinto the plane of paper That is negative \(z\) axis. \(\vec{B}=B(-\hat{k})\) Proton moves al ong \(\mathrm{x}\) axis with velocity \(\overrightarrow{\mathrm{V}}=\hat{v i}\) According to amperes right handrule That is the proton is to move normal \((\theta)=90^{\circ}\) to the \((\mathrm{B})\) Then the path foll owed by the particle is a circel

Force experienced by the proton moving with a vel ocity \((\mathrm{v})\) is

$$ \begin{aligned} F &=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}) \\ &=\mathrm{q} \mathrm{V} \mathrm{B} \sin (\theta)(\hat{i} \times-\hat{k}) \\ &=\mathrm{q} \mathrm{V} \mathrm{B} \sin \left(90^{\circ}\right)(\hat{j}) \\ &=\mathrm{q} \mathrm{VB} \hat{j} \end{aligned} $$

Direction of proton is tve y axis or north

(b)

A particle moves in a region of uniform magnetic field

with magnitude (B)(B)is out the plane of paper That is positive \(z\) axis. \(\vec{B}=B \hat{k}\) Electron moves al ong \(\mathrm{y}\) axis with velocity \(\overrightarrow{\mathrm{V}}=v \hat{j}\) According to amperes right hand rule That is the proton is to move normal \((\theta)=90^{\circ}\) to the \((\mathrm{B})\) Then the path foll owed by the particle is a circel Force experienced by the proton moving with a velocity \((\mathrm{v})\) is

$$ \begin{aligned} F &=-\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}) \\ &=-\mathrm{q} \mathrm{V} \mathrm{B} \sin (\theta)(\hat{j} \times \hat{k}) \\ &=-\mathrm{q} \mathrm{V} \mathrm{B} \sin \left(90^{\circ}\right)(\hat{i}) \\ &=\mathrm{q} \mathrm{VB}(\hat{-i}) \end{aligned} $$

Direction of electron is -ve \(\mathrm{x}\) axis or west