A charged particle enters in uniform magnetic at an angle $t a n^{- 1} {2 π}$ . This angle is between velocity vector and magnegtic field vector. Its path becomes helix with radius of path ‘r” and pitch of “p”, then

p = \sqrt{3} r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

p = \frac{r}{\sqrt{3}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

p = r

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

p = \frac{r}{\sqrt{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C p = r
$r = \frac{m v s i n θ}{q B}$
$p = \frac{2 π m v c o s θ}{q B}$
$\frac{p}{r} = \frac{t a n θ}{2 π} = \frac{2 π}{2 π} = 1$
p = r

Suggest Corrections

Similar questions

Q. A charged particle enters in uniform magnetic at an angle

t a n^{- 1} {2 π}

. This angle is between velocity vector and magnegtic field vector. Its path becomes helix with radius of path ‘r” and pitch of “p”, then

Q. A charged particle enters a uniform magnetic field with velocity vector at an angle of

45^{o}

with the magnetic field. The pitch of the helical path followed by the particle is

p .

The radius of the helix will be

Q. A charge particle enters a uniform magnetic field with velocity vector at angle of

45^{\circ}

with the magnetic field. The pitch of the helical path followed by the particle is

P

. The radius of the helix will be

A charged particle enters a uniform magnetic field with velocity vector at an angle of 45⁰ with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be

Q. A charged particle enters a uniform magnetic field with velocity that makes an angle of

45^{\circ}

with the magnetic field. The pitch of the helical path followed by the particle is

p .

The radius of the helix is

Join BYJU'S Learning Program

A charged particle enters in uniform magnetic at an angle tan−1{2π}. This angle is between velocity vector and magnegtic field vector. Its path becomes helix with radius of path ‘r” and pitch of “p”, then

The correct option is C p = rr=mv sin θqB p=2π m v cos θqB pr=tan θ2π=2 π2 π=1 p = r

A charged particle enters in uniform magnetic at an angle $t a n^{- 1} {2 π}$ . This angle is between velocity vector and magnegtic field vector. Its path becomes helix with radius of path ‘r” and pitch of “p”, then

The correct option is C p = r
$r = \frac{m v s i n θ}{q B}$
$p = \frac{2 π m v c o s θ}{q B}$
$\frac{p}{r} = \frac{t a n θ}{2 π} = \frac{2 π}{2 π} = 1$
p = r