Doubly ionized helium ions are projected with a speed of 10 km s^ -1 perpendicular to a uniform magnetic field of magnitude 1.0 T . Calculate the force acting on an ion.

Correct answer: C. 3.2 × 10^ -15 N The charge of a doubly ionized helium ion is q = +2e = 2 × 1.6 × 10^ -19 C = 3.2 × 10^ -19 C . The speed of the ion is v = 10 km s^ -1 = 10^4 m s^ -1 . The magnetic field is B = 1.0 T . The force acting on the ion moving perpendicular to the magnetic field is given by F = qvB . Substituting the values: F = 3.2 × 10^ -19 × 10^4 × 1.0 F = 3.2 × 10^ -15 N

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

Doubly ionized helium ions are projected with a speed of 10 km s^...

Doubly ionized helium ions are projected with a speed of

10 \text{ km s}^{-1}

perpendicular to a uniform magnetic field of magnitude

1.0 \text{ T}

. Calculate the force acting on an ion.

Options

A $3.2 \times 10^{-19} \text{ N}$
B $6.4 \times 10^{-15} \text{ N}$
C $3.2 \times 10^{-15} \text{ N}$
D $1.6 \times 10^{-15} \text{ N}$

Correct answer

C. $3.2 \times 10^{-15} \text{ N}$

Step-by-step solution

The charge of a doubly ionized helium ion is

q = +2e = 2 \times 1.6 \times 10^{-19} \text{ C} = 3.2 \times 10^{-19} \text{ C}

.

The speed of the ion is

v = 10 \text{ km s}^{-1} = 10^4 \text{ m s}^{-1}

.

The magnetic field is

B = 1.0 \text{ T}

.

The force acting on the ion moving perpendicular to the magnetic field is given by

F = qvB

.

Substituting the values:

F = 3.2 \times 10^{-19} \times 10^4 \times 1.0

F = 3.2 \times 10^{-15} \text{ N}

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