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
Grade: upto college level
        A particle having a charge of 5.0 μC and a mass of 5.0 × 10^-12 kg is projected with a speed of 1.0 km/s in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin^-1 (0.90) . Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.
5 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							Sol. q = 5 μF = 5 × 10^–6 C, m = 5 × 10^–12 kg, V = 1 km/s = 10^3 m/’
⇒ = Sin^–1 (0.9), B = 5 × 10^–3 T
We have mv’2 = qv’B r = mv’/qB = mv sin θ/qB = 5 * 10^-12 *10^3 * 9/5 * 10^-6 + 5 * 10^3 + 10 = 0.18 metre
Hence dimeter = 36 cm.,
Pitch = 2πr/v sin θ vcos θ = 2 * 3.1416 * 0.1 * √1 – 0.51/0.9 = 0.54 metre = 54 mc.
The velocity has a x-component along with which no force acts so that the particle, moves with uniform velocity.
The velocity has a y-component with which is accelerates with acceleration a. with the Vertical
component it moves in a circular crosssection. Thus it moves in a helix.

						
5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

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


Ask Experts

Have any Question? Ask Experts

Post Question

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