A beam of protons with speed 4 × 105 ms -1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton =1.67 × 10-27 kg, charge of the proton =1.69 × 10-19 C

Question

A beam of protons with speed 4 × 105 ms -1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton =1.67 × 10-27 kg, charge of the proton =1.69 × 10-19 C (A) 12 cm (B) 4 cm (C) 5 cm (D) 2 cm

Tardigrade Admin · Accepted Answer

Pitch =(2 π m / qB ) v cos θ Pitch =(2(3.14)(1.67 × 10-27) × 4 × 105 × cos 60/(1.69 × 10-19)(0.3)) Pitch =0.04 m =4 cm

$12 c m$

$4 c m$

$5 c m$

$2 c m$

Pitch $= \frac{2 πm}{qB} v cos θ$
Pitch $= \frac{2 ( 3.14 ) ( 1.67 \times 1 0 ^{- 27} ) \times 4 \times 1 0 ^{5} \times c o s 60}{( 1.69 \times 1 0 ^{- 19} ) ( 0.3 )}$
Pitch $= 0.04 m = 4 c m$

Q. A beam of protons with speed 4×105ms−1 enters a uniform magnetic field of 0.3T at an angle of 60∘ to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton =1.67×10−27kg, charge of the proton =1.69×10−19C

12cm

4cm

5cm

2cm

Pitch =qB2πm​vcosθ Pitch =(1.69×10−19)(0.3)2(3.14)(1.67×10−27)×4×105×cos60​ Pitch =0.04m=4cm

$12 c m$

$4 c m$

$5 c m$

$2 c m$

Pitch $= \frac{2 πm}{qB} v cos θ$
Pitch $= \frac{2 ( 3.14 ) ( 1.67 \times 1 0 ^{- 27} ) \times 4 \times 1 0 ^{5} \times c o s 60}{( 1.69 \times 1 0 ^{- 19} ) ( 0.3 )}$
Pitch $= 0.04 m = 4 c m$