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
Menu
pradyot mayank Grade: 12
        

a proton of charge e and mass m enters a uniform magnetic field B=bî with an initial velocity v=vxî+vyj(cap).find an expression in unit-vector notation for its velocity at time t. 

7 years ago

Answers : (1)

Askiitians Expert Bharath-IITD
23 Points
										Dear Padyot,


We know the formula for the Force experienced by a charged particle moving with some velocity in a magnetic field

As F = q{ V cross B )
Where V and B are in vector notations.
after Calculations we get F = {q * [(-Vy)*b]} k
Where k is a unit vector in z direction.
And the acceleration in z direction is given by

a= (Force in z direction)/mass of the charge = (q/m) * [(-Vy)*b]

We know the equation of motion as v = u + a*t
And the initial velocity in z direction is zero
and thus velocity after time t seconds in z direction is given by
Vz = a*t
Thus the velocity vector after time t seconds is given by

V = Vx i + Vy j + Vz k
= Vx i + Vy j - { (q/m) * (Vy)*b } k

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.


All the best STUDENT NAME !!!




Regards,

Askiitians Experts
Adapa Bharath
7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details