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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2025 · 7 April · Morning Shift · Physics · Magnetic Effects of Current

A particle of charge q , mass m and kinetic energy E enters in...

A particle of charge qq, mass mm and kinetic energy E enters in magnetic field perpendicular to its velocity and undergoes a circular arc of radius( r). Which of the following curves represents the variation of rr with EE ?A particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity andA particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity andA particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity andA particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity andA particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity andA particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity and

Options

  1. AThe number of critical points of the function f(x) = cases |( x)/(x)|, & x ≠ 0 1, & x = 0 cases in the interval (-2π,
  2. BThe number of critical points of the function f(x) = cases |( x)/(x)|, & x ≠ 0 1, & x = 0 cases in the interval (-2π,
  3. CThe number of critical points of the function f(x) = cases |( x)/(x)|, & x ≠ 0 1, & x = 0 cases in the interval (-2π,
  4. DThe number of critical points of the function f(x) = cases |( x)/(x)|, & x ≠ 0 1, & x = 0 cases in the interval (-2π,

Correct answer

D. The number of critical points of the function f(x) = cases |( x)/(x)|, & x ≠ 0 1, & x = 0 cases in the interval (-2π,

Step-by-step solution

A particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity and
mv2r=qvB\frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{qvB}
mv=qBr\mathrm{mv}=\mathrm{qBr}
E=12mv2\mathrm{E}=\frac{1}{2} \mathrm{mv}^2
E=12 m(q2 B2r2 m2)=q2 B2r22 m\mathrm{E}=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{q}^2 \mathrm{~B}^2 r^2}{\mathrm{~m}^2}\right)=\frac{\mathrm{q}^2 \mathrm{~B}^2 \mathrm{r}^2}{2 \mathrm{~m}}
E=(q2 B22 m)r2\mathrm{E}=\left(\frac{\mathrm{q}^2 \mathrm{~B}^2}{2 \mathrm{~m}}\right) \mathrm{r}^2
r2∝Er^2 \propto EA particle of charge q , mass m and kinetic energy E enters in magnetic field perpendicular to its velocity and
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