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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?

Vatsala , 9 Years ago
Grade 12th pass
anser 3 Answers
Divya

Last Activity: 7 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

Last Activity: 7 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

Last Activity: 5 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

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya
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