A charge particle of 2 μ C accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4 mT at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm inside magnetic field. The mass of the charge particle is × 10-18 kg

Question

A charge particle of 2 μ C accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4 mT at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm inside magnetic field. The mass of the charge particle is × 10-18 kg (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

r=(m v/q B)=(√2 k m/q B), m=(r2 q2 B2/2 k) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/016f1dedd4b7b1c8c8765e1921e29b30-.png" /> m =((1/100) × (3/100) × 2 × 2 × 4 × 10-3 × 4 × 10-3 × 10-12/2 ×(100) × 10-6) =144 × 10-18 kg

r=qBmv​=qB2km​​,m=2kr2q2B2​ m=2×(100)×10−61001​×1003​×2×2×4×10−3×4×10−3×10−12​ =144×10−18kg

$r = \frac{m v}{qB} = \frac{2 km}{qB}, m = \frac{r ^{2} q ^{2} B ^{2}}{2 k}$

$m = \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 1 0 ^{- 3} \times 4 \times 1 0 ^{- 3} \times 1 0 ^{- 12}}{2 \times ( 100 ) \times 1 0 ^{- 6}}$
$= 144 \times 1 0^{- 18} k g$