A proton accelerated by a potential difference V, moves through a transverse magnetic field B as shown in the figure. Then the angle θ through which the proton deviates from the initial direction of its motion is

Question

A proton accelerated by a potential difference V, moves through a transverse magnetic field B as shown in the figure. Then the angle θ through which the proton deviates from the initial direction of its motion is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-dgbze3qixqiu.png" /> (A) θ= sin -1(d B √(q V/2 m)) (B) θ= sin -1(d B √(q/2 m V)) (C) θ= tan -1(d B √(q V/2 m)) (D) θ= tan -1(d B √(q/2 m V))

Tardigrade Admin · Accepted Answer

B q v=(m v2/r) (1/2) m v2=V q And sin θ=(d/r) r=(m v/q B)=(m/q B) √(2 V q/m) sin θ=d B √(q/2 m V) θ= sin -1(d B √(q/2 m V))

Q. A proton accelerated by a potential difference $V,$ moves through a transverse magnetic field $B$ as shown in the figure. Then the angle $θ$ through which the proton deviates from the initial direction of its motion is

$θ = sin^{- 1} (d B \frac{q V}{2 m})$

$θ = sin^{- 1} (d B \frac{q}{2 mV})$

$θ = tan^{- 1} (d B \frac{q V}{2 m})$

$θ = tan^{- 1} (d B \frac{q}{2 mV})$

$Bq v = \frac{m v ^{2}}{r}$
$\frac{1}{2} m v^{2} = V q$
And $sin θ = \frac{d}{r}$
$r = \frac{m v}{qB} = \frac{m}{qB} \frac{2 V q}{m}$
$sin θ = d B \frac{q}{2 mV}$
$θ = sin^{- 1} (d B \frac{q}{2 mV})$

Q. A proton accelerated by a potential difference V, moves through a transverse magnetic field B as shown in the figure. Then the angle θ through which the proton deviates from the initial direction of its motion is

θ=sin−1(dB2mqV​​)

θ=sin−1(dB2mVq​​)

θ=tan−1(dB2mqV​​)

θ=tan−1(dB2mVq​​)