A charge having elm equal to 108 C/kg and with velocity 3 × 10 5 m/s enters into a uniform magnetic field B = 0.3 tesla at an angle 30° with direction of field. The radius of curvature will be

Question

A charge having elm equal to 108 C/kg and with velocity 3 × 10 5 m/s enters into a uniform magnetic field B = 0.3 tesla at an angle 30° with direction of field. The radius of curvature will be (A) 0.01 cm (B) 0.5 cm (C) 1 cm (D) 2 cm

Tardigrade Admin · Accepted Answer

r =( mV perp/ qB ) r =|( m / q )||(3 × 105 × sin 30°/0.3)| r =(3 × 105/108 × 0.3 × 2)=0.5 × 10-2 m =0.5 cm

Q. A charge having elm equal to $1 0^{8} C / k g$ and with velocity $3 \times 1 0^{5} m / s$ enters into a uniform magnetic field B = 0.3 tesla at an angle 30 $^{\circ}$ with direction of field. The radius of curvature will be

0.01 cm

0.5 cm

1 cm

2 cm

$r = \frac{m V _{⊥}}{qB}$
$r = ∣ ∣ \frac{m}{q} ∣ ∣ ∣ ∣ \frac{3 \times 1 0 ^{5} \times s i n 3 0 ^{\circ}}{0.3} ∣ ∣$
$r = \frac{3 \times 105}{1 0 ^{8} \times 0.3 \times 2} = 0.5 \times 1 0^{- 2} m = 0.5 c m$

Q. A charge having elm equal to 108C/kg and with velocity 3×105m/s enters into a uniform magnetic field B = 0.3 tesla at an angle 30∘ with direction of field. The radius of curvature will be