 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
proton move rectilinearly in the space region where there are uniform mutually perpendicular electric and magnetic fields E and B. The trajectory of the protons lie in the plane xz and forms an angle `y` with x axis. Find the pitch of the helical trajectory along which the protons will move after the elecric field is switched off.

```
11 years ago

```							(Note: initial speed is also needed to solve the problem)
Since E is prependicular to B
one can write the velocity of the proton as
V = alpha(t)*E+beta(t)*(ExB)+gamma(t)*B
where E,(ExB),B constitue the mutually orthogonal axes.

Equation of motion of the proton in
m*dV/dt=q*(E+VxB)=q*[(1-beta*|B|^2)*E+alpha*(ExB)].------(1)
which leads to following equations for the alpha(t),beta(t),gamma(t)
a) gamma(t)=const.=gamma.
b) m*d(alpha)/dt=q*(1-beta*|B|^2).
c) m*d(beta)/dt=q*alpha.
Given that the initial motion of the Proton in rectilinear
we must have alpha=const., beta=const.. On inspection of b) and c)
we obtain beta = 1/|B|^2 and alpha=0.

Hence before E is switched off: V_0=(ExB)/|B|^2+gamma*B, where gamma is a constant(can be determined from the condition given in the question before switching off E).

When E is switched off: keeping E,(ExB),B as the axes
so that V = alpha(t)*E+beta(t)*(ExB)+gamma*B
m*dV/dt=q*(VxB)=q*[-beta(t)*|B|^2*E+alpha(t)*(ExB)],
b) m*d(alpha)/dt=-q*beta*|B|^2.
c) m*d(beta)/dt=q*alpha.
elimanating alpha we get
d^2(beta)/dt=-(q*|B|/m)^2*beta -->beta(t)=A1*cos(w*t)+A2*sin(w*t),w=(q*|B|/m),
and alpha(t)=-A1*|B|*sin(w*t)+A2*|B|*cos(w*t), where A1 and A2 are constants.
Using the initial condition V_0 we have
alpha(0)=0 ==> A2=0.
beta(0)=1/|B|^2 ==>A1=1/|B|^2.

Hence
V = {cos(w*t)/|B|^2}E+{-sin(wt)/|B|}*(ExB)+gamma*B
position vector works out to be
R = {cos(w*t)/(w*|B|^2)}E+{-sin(wt)/(w*|B|)}*(ExB)+gamma*t*B
let us call
X(t)=|E|*cos(w*t)/(w*|B|^2), Y(t)=-|E|*sin(w*t)/w, Z(t)=gamma*t*|B|

the base of the helix is: (X/|B|)^2+Y^2=(|E|/w)^2;
the pitch H is given by taking t=2*pi/w
H=Z(2*pi/w)=2*pi*gamma*|B|/w=2*pi*m*gamma/q

if the proton makes an angle "theta" with the direction of B
|B|*gamma =|V_0|cos(theta)
and H = 2*pi*|V_0|*cos(theta)/w.

```
11 years ago
```							(Note: because of the confusion of notation in the question we take the angle to be "theta" between the trajectory of the proton and "B" field before switching off the "E" field).
-----------------------------------------------------------------------------------------------------------------------------------------------
Since E is prependicular to B
one can write the velocity of the proton as
V = alpha(t)*E+beta(t)*(ExB)+gamma(t)*B
where E,(ExB),B constitue the mutually orthogonal axes.

Equation of motion of the proton in
m*dV/dt=q*(E+VxB)=q*[(1-beta*|B|^2)*E+alpha*(ExB)].------(1)
which leads to following equations for the alpha(t),beta(t),gamma(t)
a) gamma(t)=const.=gamma.
b) m*d(alpha)/dt=q*(1-beta*|B|^2).
c) m*d(beta)/dt=q*alpha.
Given that the initial motion of the Proton in rectilinear
we must have alpha=const., beta=const.. On inspection of b) and c)
we obtain beta = 1/|B|^2 and alpha=0.

Hence before E is switched off: V_0=(ExB)/|B|^2+gamma*B, where gamma is a constant(can be determined from the condition given in the question before switching off E).

When E is switched off: keeping E,(ExB),B as the axes
so that V = alpha(t)*E+beta(t)*(ExB)+gamma*B
m*dV/dt=q*(VxB)=q*[-beta(t)*|B|^2*E+alpha(t)*(ExB)],
b) m*d(alpha)/dt=-q*beta*|B|^2.
c) m*d(beta)/dt=q*alpha.
elimanating alpha we get
d^2(beta)/dt=-(q*|B|/m)^2*beta -->beta(t)=A1*cos(w*t)+A2*sin(w*t),w=(q*|B|/m),
and alpha(t)=-A1*|B|*sin(w*t)+A2*|B|*cos(w*t), where A1 and A2 are constants.
Using the initial condition V_0 we have
alpha(0)=0 ==> A2=0.
beta(0)=1/|B|^2 ==>A1=1/|B|^2.

Hence
V = {cos(w*t)/|B|^2}E+{-sin(wt)/|B|}*(ExB)+gamma*B
position vector works out to be
R = {cos(w*t)/(w*|B|^2)}E+{-sin(wt)/(w*|B|)}*(ExB)+gamma*t*B
let us call
X(t)=|E|*cos(w*t)/(w*|B|^2), Y(t)=-|E|*sin(w*t)/w, Z(t)=gamma*t*|B|

the base of the helix is: (X/|B|)^2+Y^2=(|E|/w)^2;
the pitch H is given by taking t=2*pi/w
H=Z(2*pi/w)=2*pi*gamma*|B|/w=2*pi*m*gamma/q

if the proton makes an angle "theta" with the direction of B
|B|*gamma =|V_0|*cos(theta) and |V_0|*sin(theta) =|E|/|B|
so that |B|*gamma = |E|*cot(theta)/|B|
and H = 2*pi*|E|*cot(theta)/(w*|B|).

```
11 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Magnetism

View all Questions »  ### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions