Question

the path of charged particle entering in a uniform magnetic field at a angle of 60 degree is ​

Answer 1

Answer:

Pitch of the helix is product of velocity component along the magnetic field and the time period T.

Pitch: p=

qB

2πm

vcosθ=

qB

2πPcosθ

=

qB

2πPcos45

0

Here P denotes momentum

Radius of the helix: r=

qB

mvsinθ

=

qB

mvsin45

0

=

qB

Psin45

0

∴p=2πr

⇒r=

2π

p

Answer 2

Answer:

The path of charged particle entering in a uniform magnetic field is $r=\frac{\sqrt{3} x}{2 \pi}$

Explanation:

Pitch

The frequency of sound waves, which are any compression waves in a medium, is comparable to pitch.

$\begin{aligned}&P_{1}=\frac{2 \pi m}{q B} \cdot v \cos \theta \\&P_{P}=\frac{2 \pi(m v)}{q B} \cos \theta \\&\frac{2 \pi m v}{q B} \times \cos 60 \Rightarrow \frac{\pi m v}{q B} \\&P_{P}=\frac{\pi m v}{q B} \Rightarrow \frac{m v}{q B}=\frac{P_{i+c l}}{\pi} \text { }\end{aligned}$

Now radius,

 $\gamma=\frac{m v \sin \theta}{q \beta}$

$\begin{aligned}r &=\frac{m v}{2 B} \times \sin 60 \\r &=\frac{m v}{q B} \times \frac{\sqrt{3}}{2} \\r &=\frac{\text { pitch }}{\pi} \times \frac{\sqrt{3}}{2} \Rightarrow \frac{\sqrt{3} x}{2 \pi} \\r=\frac{\sqrt{3} x}{2 \pi}\end{aligned}$

Thus, the charged path is  $r=\frac{\sqrt{3} x}{2 \pi}$.