JEE Main 2014 · 12 Apr · Shift 1 · Physics Q19

← JEE Main 2014 · 12 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 19 of 90

A positive charge 'q' of mass 'm' is moving along the $+x$ axis. We wish to apply a uniform magnetic field B for time Δt so that the charge reverses its direction crossing the $y$ axis at a distance d. Then :

B=\dfrac{mv}{qd}

and

\Delta t=\dfrac{\pi d}v

B=\dfrac{mv}{2qd}

and

\Delta t=\dfrac{\pi d}{2v}

B=\dfrac{2mv}{qd}

and

\Delta t=\dfrac{\pi d}{2v}

B=\dfrac{2mv}{qd}

and

\Delta t=\dfrac{\pi d}v

Answer and worked solution

Official answer: C

The charge must describe a semicircle of diameter $d$ , so $r=\frac d2=\dfrac{mv}{qB}$ gives $B=\dfrac{2mv}{qd}$ .

Half a circle takes $\Delta t=\dfrac{\pi r}v=\dfrac{\pi d}{2v}$ .

Topics:Force on a moving charge

← Previous question Next question →

JEE Main 2014 · 12 Apr · Shift 1, Physics question 19