JEE Advanced 2019 Paper 1 · Physics · Electrostatics · hard

A proton enters a region of uniform magnetic field vec B=0.10 hat k T and uniform electric field vec E=1000 hat k V m^-1 with velocity vec v0=(3hat…

One or more options correct

A proton enters a region of uniform magnetic field

\vec B=0.10\,\hat k\ \mathrm{T}

and uniform electric field

\vec E=1000\,\hat k\ \mathrm{V\,m^{-1}}

with velocity

\vec v_0=(3\hat i+4\hat k)\times10^5\ \mathrm{m\,s^{-1}}

. Use

m_p=1.67\times10^{-27}\ \mathrm{kg}

and

e=1.6\times10^{-19}\ \mathrm{C}

. Which of the following statements is/are correct?

A. The radius of the circular projection of the path on the $xy$ -plane is constant, about $3.1\ \mathrm{cm}$ .
B. The kinetic energy of the proton remains constant.
C. The time taken for each turn of the helix is the same.
D. The displacement along $z$ during the first turn is about $0.28\ \mathrm{m}$ .

Show answer and solution

Answer: A, C, D

The magnetic force acts only on the transverse velocity and the electric force is along

z

, so the motion separates into circular motion in the

xy

-plane with

v_\perp=3\times10^5\ \mathrm{m\,s^{-1}}

, and uniformly accelerated motion along

z

. Radius

r=mv_\perp/eB=1.67\times10^{-27}\times3\times10^5/(1.6\times10^{-19}\times0.1)=3.13\ \mathrm{cm}

, constant, so (A) is true. The electric field does work as the proton moves along

E

, so kinetic energy grows and (B) is false. The turn time is

T=2\pi m/eB=6.56\times10^{-7}\ \mathrm{s}

, independent of

E

and

v_z

, so (C) is true. In the first turn

\Delta z=v_zT+\tfrac12(eE/m)T^2=0.262+0.021=0.283\ \mathrm{m}

, so (D) is true. Correct options: (A), (C), (D).

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