A beam of ions enters normally into a uniform magnetic field of 4 × 10-2 T with velocity of 2 × 105 m / s. 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 enters normally into a uniform magnetic field of 4 × 10-2 T with velocity of 2 × 105 m / s. 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) 0.06 m (C) 0.20 m (D) 0.25 m

Tardigrade Admin · Accepted Answer

Force due to magnetic field provides the required centripetal force to the ions.

∴

Centripetal force = force due to magnetic field

\frac{m v ^{2}}{r} = e v B sin θ

Given,

θ = 9 0^{\circ}, \frac{e}{m} = 5 \times 1 0^{7} C / k g

,

B = 4 \times 1 0^{- 2} T,

v = 2 \times 1 0^{5} m / s

∴ r = \frac{m v}{e B}

\Rightarrow r = \frac{2 \times 1 0 ^{5}}{5 \times 1 0 ^{7} \times 4 \times 1 0 ^{- 2}}

= 0.1 m

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

0.10 m

0.06 m

0.20 m

0.25 m

Force due to magnetic field provides the required centripetal force to the ions. ∴ Centripetal force = force due to magnetic field rmv2​=evBsinθ Given, θ=90∘,me​=5×107C/kg, B=4×10−2T, v=2×105m/s ∴r=eBmv​ ⇒r=5×107×4×10−22×105​ =0.1m

Q. A beam of ions enters normally into a uniform magnetic field of $4 \times 1 0^{- 2} T$ with velocity of $2 \times 1 0^{5} m / s$ . 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