A particle of specific charge - starts moving from the origin under the action of an electric field E- - E0 - and magnetic field B- - B0 k- Its velocity at x0- y0- 0 is 4 - - 3 - The value of x0 is

From work energy theorem, Δ W_E + Δ W_B = Δ K ⇒ qE_0x_0+0 = 1/2 mv^2=1/2 m(3^2+4^2) ⇒q/mE_0x_0 = 25/2 ⇒α E_0x_0 = 25/2 ∴ x_0=25/2 α E_0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Helical Motion

A particle of...

Question

A particle of specific charge $α$ starts moving from the origin under the action of an electric field $\to E = E_{0}^i$ and magnetic field $\to B = B_{0}^k$ . Its velocity at $(x_{0}, y_{0}, 0)$ is $(4^i + 3^j)$ . The value of $x_{0}$ is

$\frac{16 α B_{0}}{E_{0}}$

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

$\frac{13 α E_{0}}{2 B_{0}}$

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

$\frac{25}{2 α E_{0}}$

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

$\frac{5 α}{2 B_{0}}$

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

Open in App

Solution

The correct option is C
$\frac{25}{2 α E_{0}}$

From work-energy theorem, $Δ W_{E} + Δ W_{B} = Δ K$
$\Rightarrow q E_{0} x_{0} + 0 = \frac{1}{2} m v^{2} = \frac{1}{2} m (3^{2} + 4^{2})$
$\Rightarrow \frac{q}{m} E_{0} x_{0} = \frac{25}{2}$
$\Rightarrow α E_{0} x_{0} = \frac{25}{2}$
$∴ x_{0} = \frac{25}{2 α E_{0}}$

Suggest Corrections

Similar questions

Q. A particle of specific charge (charge/mass) starts moving from the origin under the action of an electric field

{E = E}_{0}^i

and magnetic field

B = B_{0}^k

Its velocity at

(x_{0}, y_{0}, 0)

(4^i - 3^j)

The value of

x_{0}

A particle of mass m and charge q starts moving from the origin under the action of an electric field $\to E = E_{0}^i$ and magnetic field $\to B = B_{0}^k$ . Its velocity at $(x_{0}, 0, 0)$ is $(4^i + 3^j)$ . Then

Q. A particle of specific charge (charge/mass)

α

starts moving from the origin from rest under the action of an electric field

\to E = E_{0}^i

and magnetic field

\to B = B_{0}^k

. Its velocity at

(x_{0}, y_{0}, 0)

(4^i - 3^j)

. The value of

x_{0}

is :
(All parameters are in SI unit.)

Q. A particle of charge

q

and mass

m

starts moving from origin under the action of an electric field

\to E

E_{0} \to i

and magnetic field

\to B

B_{0} \to k

. Its velocity at

(x, 3, 0)

(4 i + 3 i)

, the value of

x

is:

Q. A particle of charge q and mass m starts moving from the origin under the action of an electric field

\to E = E_{0}^i

and

\to B = B_{0}^i

with a velocity

\to v = v_{0}^i

. The speed of the particle will become

2 v_{0}

after a time.

Join BYJU'S Learning Program

A particle of specific charge α starts moving from the origin under the action of an electric field →E=E0^i and magnetic field →B=B0^k. Its velocity at (x0,y0,0) is (4^i+3^j). The value of x0 is

The correct option is C 252αE0 From work-energy theorem, ΔWE+ΔWB=ΔK ⇒qE0x0+0=12mv2=12m(32+42) ⇒qmE0x0=252 ⇒αE0x0=252 ∴x0=252αE0

A particle of specific charge $α$ starts moving from the origin under the action of an electric field $\to E = E_{0}^i$ and magnetic field $\to B = B_{0}^k$ . Its velocity at $(x_{0}, y_{0}, 0)$ is $(4^i + 3^j)$ . The value of $x_{0}$ is