A particle with a mass of 2.0 × 10^ -10 g and a charge of 2.0 × 10^ -8 C is projected at a speed of 2.0 × 10^3 m s^ -1 into a region containing a uniform magnetic field of 0.10 T . The particle's velocity is

Correct answer: A. 20 cm and 6.3 × 10^ -4 s Given mass of the particle, m = 2.0 × 10^ -10 g = 2.0 × 10^ -13 kg Charge, q = 2.0 × 10^ -8 C Speed, v = 2.0 × 10^3 m s^ -1 Magnetic field, B = 0.10 T The radius of the circular path is given by: r = (mv)/(qB) Substituting the given values: r = 2.0 × 10^ -13 × 2.0 × 10^3 2.0 × 10^ -8 × 0.10 r = 4.0 × 10^ -10 2.0 × 10^ -9 = 0.2 m = 20 cm The time period of revolution is given by: T = (2π m)/(qB) = (2π r)/(v) Substituting the values: T = (2 × 3.14 × 0.2)/(2.0 × 10^3) T = (1.256)/(2.0 × 10^3) = 6.28 × 10^ -4 s 6.3 × 10^ -4 s

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Magnetic Field

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Magnetic Field

A particle with a mass of 2.0 × 10^ -10 g and a charge of 2.0 × 10^...

A particle with a mass of

2.0 \times 10^{-10} \text{ g}

and a charge of

2.0 \times 10^{-8} \text{ C}

is projected at a speed of

2.0 \times 10^3 \text{ m s}^{-1}

into a region containing a uniform magnetic field of

0.10 \text{ T}

. The particle's velocity is perpendicular to the magnetic field. Determine the radius of the circular path described by the particle, as well as its time period.

Options

A $20 \text{ cm}$ and $6.3 \times 10^{-4} \text{ s}$
B $200 \text{ m}$ and $0.63 \text{ s}$
C $40 \text{ cm}$ and $1.3 \times 10^{-3} \text{ s}$
D $10 \text{ cm}$ and $3.1 \times 10^{-4} \text{ s}$

Correct answer

A. $20 \text{ cm}$ and $6.3 \times 10^{-4} \text{ s}$

Step-by-step solution

Given mass of the particle,

m = 2.0 \times 10^{-10} \text{ g} = 2.0 \times 10^{-13} \text{ kg}

Charge,

q = 2.0 \times 10^{-8} \text{ C}

Speed,

v = 2.0 \times 10^3 \text{ m s}^{-1}

Magnetic field,

B = 0.10 \text{ T}

The radius of the circular path is given by:

r = \dfrac{mv}{qB}

Substituting the given values:

r = \dfrac{2.0 \times 10^{-13} \times 2.0 \times 10^3}{2.0 \times 10^{-8} \times 0.10}

r = \dfrac{4.0 \times 10^{-10}}{2.0 \times 10^{-9}} = 0.2 \text{ m} = 20 \text{ cm}

The time period of revolution is given by:

T = \dfrac{2\pi m}{qB} = \dfrac{2\pi r}{v}

Substituting the values:

T = \dfrac{2 \times 3.14 \times 0.2}{2.0 \times 10^3}

T = \dfrac{1.256}{2.0 \times 10^3} = 6.28 \times 10^{-4} \text{ s} \approx 6.3 \times 10^{-4} \text{ s}

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