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

Byju's Answer

Standard XII

Physics

A Mysterious Force

A velocity se...

Question

A velocity selector consists of electric field $\to E = E^k$ and magnetic field $\to B = B^j$ with $B = 12 mT$ . The value of $E$ required for an electron of energy $728 eV$ moving along the positive $x$ -axis to pass undeflected is (Given, mass of electron $= 9.1 \times 10^{- 31} kg$ )

9600 {kVm}^{- 1}

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

192 {mVm}^{- 1}

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

192 {kVm}^{- 1}

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

16 {kVm}^{- 1}

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

Open in App

Solution

The correct option is C $192 {kVm}^{- 1}$
As we know that,

velocity of selector, $v = \frac{E}{B}$ and

Kinetic energy, $K = \frac{1}{2} m V^{2}$

$\Rightarrow \sqrt{\frac{2 K}{m}} \times B = E$

$\Rightarrow E = \sqrt{\frac{2 \times 728 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}} \times 12 \times 10^{- 3}$

$= 192000 V/m$

Hence, $(A)$ is the correct answer.

Suggest Corrections

Similar questions

An electron moving along positive X-axis with a $3 \times 10^{6}$ m/s speed, enters the region of a uniform electric field. The electron stops after travelling a distance of $90 m m$ in the field. The electric field strength is (charge on electron $= 1.6 \times 10^{- 19}$ , mass of electron $= 9.1 \times 10^{- 31} k g$ ) :

Q. In J.J.Thomson's method , electric field E, magnetic field B and velocity v of the electrons were in mutually perpendicular direction. This velocity selector allows particles of velocity v to pass undeflected when

Q. An electric field

E

and a magnetic field

B

act over the same region in which an electron enters along the x-axis. The combination of

E

and

B

which permits the electron to go undeflected is:

Q. An electron is released from the origin at a place where a uniform electric field

E

and a uniform magnetic field

B

exist along the negative

y

-axis and the negative

z

-axis respectively. Find the displacement of the electron along the

y

-axis when its velocity becomes perpendicular to the electric field for the first time. (Take

E = 20 V/m

and

B = 0.5 mT

, charge of electron

e = 1.6 \times 10^{- 19} C

, Mass of electron

m = 10^{- 30} kg

)

an electron beam passes through a_{REGION OF CROSSED ELECTRIC AND MAGNETIC FIELDS OF INTENSITIES} E B RESPECTIVELY. FOR WHAT VALUE OF THE ELECTRON SPPED WILL THE ELECTRON BEAM REMAIN UNDEFLECTED?

Join BYJU'S Learning Program

A velocity selector consists of electric field →E=E^k and magnetic field →B=B^j with B=12 mT. The value of E required for an electron of energy 728 eV moving along the positive x-axis to pass undeflected is (Given, mass of electron =9.1×10–31 kg)

The correct option is C 192 kVm−1As we know that, velocity of selector, v=EB and Kinetic energy, K=12mV2 ⇒ √2Km×B=E ⇒ E=√2×728×1.6×10−199.1×10−31×12×10−3 =192000 V/m Hence, (A) is the correct answer.

A velocity selector consists of electric field $\to E = E^k$ and magnetic field $\to B = B^j$ with $B = 12 mT$ . The value of $E$ required for an electron of energy $728 eV$ moving along the positive $x$ -axis to pass undeflected is (Given, mass of electron $= 9.1 \times 10^{- 31} kg$ )