A charged particle enters a uniform magnetic field with velocity vector at an angle of 45° with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be

Question

A charged particle enters a uniform magnetic field with velocity vector at an angle of 45° with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be (A) (p/√2 π) (B) √2 p (C) (p/2 π) (D) (√2 p/π)

Tardigrade Admin · Accepted Answer

p=(2 π m/B q)(v cos 45°)=(2 π m/B q)(v sin 45°) ∴ (m v sin 45°/B q)=(p/2 π)= radius of helix

Q. A charged particle enters a uniform magnetic field with velocity vector at an angle of $4 5^{\circ}$ with the magnetic field. The pitch of the helical path followed by the particle is $p$ . The radius of the helix will be

$\frac{p}{2 π}$

$2 p$

$\frac{p}{2 π}$

$\frac{2 p}{π}$

$p = \frac{2 πm}{Bq} (v cos 4 5^{\circ}) = \frac{2 πm}{Bq} (v sin 4 5^{\circ})$
$∴ \frac{m v s i n 4 5 ^{\circ}}{Bq} = \frac{p}{2 π} =$ radius of helix

Q. A charged particle enters a uniform magnetic field with velocity vector at an angle of 45∘ with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be

2​πp​

2​p

2πp​

π2​p​

p=Bq2πm​(vcos45∘)=Bq2πm​(vsin45∘) ∴Bqmvsin45∘​=2πp​= radius of helix

Q. A charged particle enters a uniform magnetic field with velocity vector at an angle of $4 5^{\circ}$ with the magnetic field. The pitch of the helical path followed by the particle is $p$ . The radius of the helix will be

$\frac{p}{2 π}$

$2 p$

$\frac{p}{2 π}$

$\frac{2 p}{π}$

$p = \frac{2 πm}{Bq} (v cos 4 5^{\circ}) = \frac{2 πm}{Bq} (v sin 4 5^{\circ})$
$∴ \frac{m v s i n 4 5 ^{\circ}}{Bq} = \frac{p}{2 π} =$ radius of helix