A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6 × 10-19 C)

Question

A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6 × 10-19 C) (A) 2.0 × 10-24 kg  (B) 1.6 × 10-19 kg  (C) 1.6 × 10-27 kg  (D) 9.1 × 10-31 kg

Tardigrade Admin · Accepted Answer

(mv2/R) = qvB mv =qBR ....(i) Path is straight line it qE = qvB E = vB ....(ii) From equation (i) (ii) m = (qB2R/E) m = 2.0 ×10-24 kg

$2.0 \times 1 0^{- 24} k g$

$1.6 \times 1 0^{- 19} k g$

$1.6 \times 1 0^{- 27} k g$

$9.1 \times 1 0^{- 31} k g$

$\frac{m v ^{2}}{R} = q v B$
$m v = qBR$ ....(i)
Path is straight line
it $qE = q v B$
$E = v B$ ....(ii)
From equation (i) & (ii)
$m = \frac{q B ^{2} R}{E}$
$m = 2.0 \times 1 0^{- 24} k g$

Q. A particle having the same charge as of electron moves in a circular path of radius 0.5cm under the influence of a magnetic field of 0.5T. If an electric field of 100V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6×10−19C)

2.0×10−24kg

1.6×10−19kg

1.6×10−27kg

9.1×10−31kg

Rmv2​=qvB mv=qBR ....(i) Path is straight line it qE=qvB E=vB ....(ii) From equation (i) & (ii) m=EqB2R​ m=2.0×10−24kg

$2.0 \times 1 0^{- 24} k g$

$1.6 \times 1 0^{- 19} k g$

$1.6 \times 1 0^{- 27} k g$

$9.1 \times 1 0^{- 31} k g$

$\frac{m v ^{2}}{R} = q v B$
$m v = qBR$ ....(i)
Path is straight line
it $qE = q v B$
$E = v B$ ....(ii)
From equation (i) & (ii)
$m = \frac{q B ^{2} R}{E}$
$m = 2.0 \times 1 0^{- 24} k g$