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​Q. The magnetic in a certain region of space is given by- B = 8.35 * 10-2 j^ T. A proton is shot intothe field with velocity v= ( 2 * 105 i^ + 4 * 105 j^ ) The proton follows a helical path in the field. The distance moved by proton in the x- direction during the period of one revolution in the yz- plane will be- 1. 0.053m 2. 0.136m 3. 0.157m 4. 0.236m

Naina Singh , 8 Years ago
Grade 12th pass
anser 1 Answers
deepak

Last Activity: 8 Years ago

first let us understand the motion of the particle,
 
F = q \overrightarrow{V} X\overrightarrow{B}
hence force is going to be towards the \widehat{k}
hence the particle will have circular motion in the x-z plane and translatory motion along the y direction
 
 
hence the distance moved along the x direction will be the radius of the helix,
 
we know that the radius of the helix = MV / BQ …....where m is mass of the proton q is its charge v is its velocity along the x                                                                                direction ( only the i component) and B is the magnetic field strength
=>r =  0.0236 m
 
the answer resembles option 4 pls check whether you have correctly given the information
 
hope this helps XD

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