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Grade: 12
        
A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1.0 × 10^6 m/s. It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle described by it. 
one year ago

Answers : (3)

Arun
22809 Points
							
Dear Kiran
 
V = 12 KV E = V/l Now, F = qE = qV/ld or, a = F/m = qV/ml υ = 1 * 10^6 m/s Or V = √2 * qV/ml*l = √2 * q/m * 12 * 10^3 Or 1 * 10^6 = √2 * q/m * 12 * 10^3 ⇒ 10^12 = 24 * 10^3 * q/m ⇒ m/q = 24 * 10^3/10^12 = 24 * 10^-9 r = mV/qB = 24 * 10^-9 *1 * 10^6/2 * 10^-1 = 12 * 10^-2 m = 12 cm
one year ago
Suprotim
15 Points
							
Potenital energy lost = Kinetic energy gained
So, e*V = 0.5*m*v^2  [ e = charge, V = Volt diff., m = mass;, v = velocity]
so, e = m*0.5*v^2*1/V.
 
Now , when it enters perpendicularly in a magnetic field on 0.2 T, it will experience a force ,
Force = e*v*B*sin(theta) ,
Equate this force with centripetal force (m*v^2/r) & r will be 12 cm
one year ago
Aman
34 Points
							
Equate the energy by two formulas mv^2/2=q×V here v is velocity and V is potential difference so from here obtain the ration of m/V and radius of the circle described (r)=mv/qB put in this formula and get the answer
one year ago
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