Click to Chat

1800-2000-838

+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
Amit Saxena Grade: upto college level
`        A particle having a charge of 2.0 × 10^-8 C and a mass of 2.0 × 10^-10 g is projected with a speed of 2.0 × 10^3 m/s. in a region having a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.`
3 years ago

Navjyot Kalra
654 Points
```										Sol. q = 2.0 × 10–8 C B ⃗ = 0.10 T
m = 2.0 * 10^-10 g = 2 * 10^-13 g
υ = 2.0 * 10^3 m/’
R = m υ/qB = 2 * 10^-13 *2 * 10^3/2 * 10^-8 * 10^-1 = 0.2 m = 20 cm
T = 2πm/qB = 2 * 3.14 *2 * 10^-13/2 * 10^-18 * 10^-1 = 6.28 * 10^-4 s

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

## Other Related Questions on Electromagnetic Induction

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Electricity and Magnetism
• OFFERED PRICE: Rs. 1,696
• View Details