0

Guest

Courses New

A proton goes undeflected in a crossed in a crossed electric and magnetic field (the field are perpendicular to each other) at a speed of 2.0 × 10^5 m/s. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of the proton = 1.6 × 10^-27 kg.

A proton goes undeflected in a crossed in a crossed electric and magnetic field (the field are perpendicular to each other) at a speed of 2.0 × 10^5 m/s. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of the proton = 1.6 × 10^-27 kg.

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Sol. M base P = 1.6 × 10^–27 Kg υ = 2 × 105 m/s r = 4 cm = 4 × 10^–2 m Since the proton is undeflected in the combined magnetic and electric field. Hence force due to both the fields must be same. i.e. qE = qυB ⇒ E = υB Won, when the electricfield is stopped, then if forms a circle due to force of magnetic field We know r = mυ/qB ⇒ 4 * 10^2 = 1.6 *10^-27 *2 * 10^5/1.6 * 10^-19 *B ⇒ B = 1.6 *10^-27 *2 * 10^5/4 * 10^2 * 1.6 *10^-19 = 0.5 * 10^-1 = 0.005 T E = υB = 2 * 10^5 * 0.05 = 1 * 10^4 N/C

Think You Can Provide A Better Answer ?

Other Related Questions on Electromagnetic Induction

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More