A beam of ions with velocity 2 × 105 m / s enters normally into a uniform magnetic field of 4 × 10-2 tesla. If the specific charge of the ion is 5 × 107 C / kg then the radius of the circular path described will be

Question

A beam of ions with velocity 2 × 105 m / s enters normally into a uniform magnetic field of 4 × 10-2 tesla. If the specific charge of the ion is 5 × 107 C / kg then the radius of the circular path described will be (A) 0.10 m (B) 0.16 m (C) 0.20 m (D) 0.25 m

Tardigrade Admin · Accepted Answer

r =( mv / Bq )=( v /( q ) m × B )=(2 × 105/5 × 107 × 4 × 10-2)=0.1 m r=0.1 m

Q. A beam of ions with velocity $2 \times 1 0^{5} m / s$ enters normally into a uniform magnetic field of $4 \times 1 0^{- 2}$ tesla. If the specific charge of the ion is $5 \times 1 0^{7} C / k g$ then the radius of the circular path described will be

$0.10 m$

$0.16 m$

$0.20 m$

$0.25 m$

$r = \frac{m v}{Bq} = \frac{v}{\frac{q}{m} \times B} = \frac{2 \times 1 0 ^{5}}{5 \times 1 0 ^{7} \times 4 \times 1 0 ^{- 2}} = 0.1 m$
$r = 0.1 m$

Q. A beam of ions with velocity 2×105m/s enters normally into a uniform magnetic field of 4×10−2 tesla. If the specific charge of the ion is 5×107C/kg then the radius of the circular path described will be

0.10m

0.16m

0.20m

0.25m