A proton travels along a circle of radius 1 m in a perpendicular magnetic field of strength 0.50 T . What would its speed be?

Correct answer: D. 4.8 × 10^7 m s^ -1 The radius of the circular path of a charged particle in a perpendicular magnetic field is given by r = (mv)/(qB) Rearranging for speed v , we get v = (qBr)/(m) For a proton, charge q = 1.6 × 10^ -19 C and mass m = 1.67 × 10^ -27 kg Given B = 0.50 T and r = 1 m Substituting the values: v = 1.6 × 10^ -19 × 0.50 × 1 1.67 × 10^ -27 v = 0.8 × 10^ -19 1.67 × 10^ -27 v 4.8 × 10^7 m s^ -1

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 proton travels along a circle of radius 1 m in a perpendicular...

A proton travels along a circle of radius

1 \text{ m}

in a perpendicular magnetic field of strength

0.50 \text{ T}

. What would its speed be?

Options

A $9.6 \times 10^7 \text{ m s}^{-1}$
B $4.8 \times 10^5 \text{ m s}^{-1}$
C $2.4 \times 10^7 \text{ m s}^{-1}$
D $4.8 \times 10^7 \text{ m s}^{-1}$

Correct answer

D. $4.8 \times 10^7 \text{ m s}^{-1}$

Step-by-step solution

The radius of the circular path of a charged particle in a perpendicular magnetic field is given by

r = \dfrac{mv}{qB}

Rearranging for speed

v

, we get

v = \dfrac{qBr}{m}

For a proton, charge

q = 1.6 \times 10^{-19} \text{ C}

and mass

m = 1.67 \times 10^{-27} \text{ kg}

Given

B = 0.50 \text{ T}

and

r = 1 \text{ m}

Substituting the values:

v = \dfrac{1.6 \times 10^{-19} \times 0.50 \times 1}{1.67 \times 10^{-27}}

v = \dfrac{0.8 \times 10^{-19}}{1.67 \times 10^{-27}}

v \approx 4.8 \times 10^7 \text{ m s}^{-1}

Practice Magnetic Field on Quantrex Academy →

More from Magnetic Field