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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

deepak
84 Points
4 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