# A particle having a charge of 10 micro coulomb and a mass of1Micro gram moves in a horizontal circle of radius 10cm under the influence of a magnetic field Of0.1T.when the particle is at a Point P,a uniform electric field is switched on so that the particle start moving along the tangent with uniform velocity.What is the electric field?

Divya
32 Points
4 years ago
For the particle to go undeflected,
$qvBsin\theta =qE$
E=vB---------- (1)
mv2/r=qvb
v=qBr/m-------------(2)
From 1 and 2,
E=vB        $E=qB^2r/m$
putting the values we get,E=0.01V/m
Shubhangi Sharma
13 Points
4 years ago
qvBsin∅ = qEE = vBv = qBr/mE = qB²r/mConverting all values into standard, C, kg & m;E = (10×10^-6×0.01×0.1)÷(10^-6×1×10^-3)E = 10 V/m
Yash Chourasiya
2 years ago
Dear Student

As per condition
qVB = qE
Now,R = mv/qB
​⇒V = BqR​/M
(BqR/M)*B = E........(1)
After putting values in equation (1), we get
E = 10V