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An electron is accelerated by a potential difference of 12000 volt. It then enters a uniform magnetic field of 10 ^-3 T applied perpendicular to the path of electron .find the radius of the path. Given mass of electron is equal to 9 * 10^-31 kg and charge of electron =1.6*10^-19 C


2 years ago

Susmita
425 Points
							Say after acceleration final velocity of electron is v.V=12000volt is the potential difference.So kinetic energy of electron is $\frac{1}{2} mv^2 = eV$$Or, v= \sqrt {2eV/m}$Or,v=4.26×1015 m/sForce on a charged particle in a magnetic field B isq(vector v×B)=qvBsin(theta)=evBsin90°=evB. In case velocity is perpendicular to magnetic field,the particle will exhibit circular motion.If r is radius of circular orbit,then centripetal force $\frac{mv^2}{r} = evB$$Or, r =\sqrt{\frac{mv}{eB}}$r=24×105 m.

2 years ago
Susmita
425 Points
							It will not be root over in the last line.It was typing mistake.It will be justr=mv/eB.Please approve the previous answer if you are helped.

2 years ago
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### Course Features

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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions