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?

Question

Divya · Answer

For the particle to go undeflected, E=vB ---------- (1) mv 2 /r=qvb v=qBr/m -------------(2) From 1 and 2, E=vB putting the values we get,E=0.01V/m

Shubhangi Sharma · Answer

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 · Answer

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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