A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If α be the angle of deviation of proton from initial direction of motion (see figure), the value of sin α will be :

Question

A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If α be the angle of deviation of proton from initial direction of motion (see figure), the value of sin α will be : (A) (B/2)√(qd/mV) (B) (B/d)√(q/2mV) (C) Bd√(q/2mV) (D) qV√(Bd/2m)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fee932a27c4fadebbff4ae87a0f16ca2-.png" /> q V=(1/2) m v2 ⇒ v=√(2 q v/m) r=(m v/q B)=(m/q B) √(2 q V/m) r=(1/B) √(2 m V/q) sin α =(d/r)=B d √(q/2 m V)

Q. A proton (mass m) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$ . The field occupies a region of space by width $d$ . If $α$ be the angle of deviation of proton from initial direction of motion (see figure), the value of $sin α$ will be :

$\frac{B}{2} \frac{q d}{mV}$

$\frac{B}{d} \frac{q}{2 mV}$

$B d \frac{q}{2 mV}$

$q V \frac{B d}{2 m}$

$q V = \frac{1}{2} m v^{2}$
$\Rightarrow v = \frac{2 q v}{m} r = \frac{m v}{qB} = \frac{m}{qB} \frac{2 q V}{m}$
$r = \frac{1}{B} \frac{2 mV}{q} sin α$
$= \frac{d}{r} = B d \frac{q}{2 mV}$

Q. A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If α be the angle of deviation of proton from initial direction of motion (see figure), the value of sinα will be :

2B​mVqd​​

dB​2mVq​​

Bd2mVq​​

qV2mBd​​