A positive charge - q- of mass - m- is moving along the -x axis- We wish to apply a uniform magnetic field B for time - t- so that the charge reverses its direction- crossing the y axis at a distance d- Then-

The correct option is C B=2mvqd and Δ t=π d2v To #160;change #160;the #160;direction #160;of #160;charge, #160;a #160;force #160;nee ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Comparing Vectors

A positive ch...

Question

A positive charge ' $q$ ' of mass ' $m$ ' is moving along the +x axis. We wish to apply a uniform magnetic field $B$ for time $Δ t$ , so that the charge reverses its direction, crossing the y axis at a distance $d$ . Then:

B = \frac{m v}{q d}

and

Δ t = \frac{π d}{v}

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

B = \frac{m v}{2 q d}

and

Δ t = \frac{π d}{2 v}

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

B = \frac{2 m v}{q d}

and

Δ t = \frac{π d}{2 v}

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

B = \frac{2 m v}{q d}

and

Δ t = \frac{π d}{v}

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

Open in App

Solution

The correct option is C $B = \frac{2 m v}{q d}$ and $Δ t = \frac{π d}{2 v}$
To change the direction of charge, a force needs to be applied such that,
$q v B = \frac{m v^{2}}{r}$ where $r = d / 2$
Therefore,
$B = \frac{2 m v}{q d}$
Total displacement is $d$ when charge travels for time, $δ t$
Therefore,
$δ t = \frac{π d}{2 v}$

Suggest Corrections

Similar questions

A particle of charge q and mass m moving with velocity v along the x - axis enters the region x>0 with uniform magnetic field B along positive z direction. The particle will penetrate in this region along x - axis up to a distance d equal to

Statement-I :

The value of the integral

π / 3 \int π / 6 \frac{d x}{1 + \sqrt{tan x}}

is equal to

\frac{π}{6}

Statement-II : b \int a f (x) d (x) = b \int a f (a + b - x) d (x)

Q. Spherical CapacitorIt consists of two concentric conducting spherical shells of radii

a

and

b (a < b)

. The inner shell is given a charge

Q

and outer shell is given a charge

Q

. The potential difference between the shells is given by

V_{i} - V_{0} = - \int_{b}^{a} E d r

\Rightarrow V = - \int_{b}^{a} \frac{Q}{4 π ε_{o} r^{2}} d r

\Rightarrow V = \frac{Q (b - a)}{4 π ε_{o} a b}

Now,

C = \frac{Q}{V} \Rightarrow C = \frac{4 π ε_{o} (a b)}{(b - a)}

Q. A positive charge particle with charge

q

and mass

m

is moving with velocity

v = v_{0}^i

in a uniform magnetic field

\to B = - B_{0}^k

. The time spent by the particle in the magnetic field if

d

is equal to

\frac{m v}{2 q B}

will be.

Join BYJU'S Learning Program

A positive charge 'q' of mass 'm' is moving along the +x axis. We wish to apply a uniform magnetic field B for time Δt, so that the charge reverses its direction, crossing the y axis at a distance d. Then:

The correct option is C B=2mvqd and Δt=πd2vTo change the direction of charge, a force needs to be applied such that,qvB=mv2r where r=d/2Therefore,B=2mvqdTotal displacement is d when charge travels for time, δtTherefore,δt=πd2v

A positive charge ' $q$ ' of mass ' $m$ ' is moving along the +x axis. We wish to apply a uniform magnetic field $B$ for time $Δ t$ , so that the charge reverses its direction, crossing the y axis at a distance $d$ . Then: