A proton accelerated by a potential difference 500 kV moves though a transverse magnetic field of 0.51 T as shown in figure. The angle θ through which the proton deviates from the initial direction of its motion is

Question

A proton accelerated by a potential difference 500 kV moves though a transverse magnetic field of 0.51 T as shown in figure. The angle θ through which the proton deviates from the initial direction of its motion is (A) 15° (B) 30° (C) 45° (D) 60°

Tardigrade Admin · Accepted Answer

According to following figure sin θ=(d/r) also r=(√2 m k/q B)=(1/B) √(2 m V/q) ∴ sin θ=B d √(q/2 m V) =0.51 × 0.1 √(1.6 × 10-19/2 × 1.67 × 10-27 × 500 × 103)=(1/2) ⇒ θ=30°

Q. A proton accelerated by a potential difference $500 kV$ moves though a transverse magnetic field of $0.51 T$ as shown in figure. The angle $θ$ through which the proton deviates from the initial direction of its motion is

$1 5^{\circ}$

$3 0^{\circ}$

$4 5^{\circ}$

$6 0^{\circ}$

According to following figure
$sin θ = \frac{d}{r}$
also $r = \frac{2 mk}{qB} = \frac{1}{B} \frac{2 mV}{q}$
$∴ sin θ = B d \frac{q}{2 mV}$
$= 0.51 \times 0.1 \frac{1.6 \times 1 0 ^{- 19}}{2 \times 1.67 \times 1 0 ^{- 27} \times 500 \times 1 0 ^{3}} = \frac{1}{2}$
$\Rightarrow θ = 3 0^{\circ}$

Q. A proton accelerated by a potential difference 500kV moves though a transverse magnetic field of 0.51T as shown in figure. The angle θ through which the proton deviates from the initial direction of its motion is

15∘

30∘

45∘

60∘