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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

In a region where a 1.0 T magnetic field points from south to...

In a region where a 1.0 T1.0 \text{ T} magnetic field points from south to north, an alpha particle is projected vertically upward at a speed of 3.0×104 km s−13.0 \times 10^4 \text{ km s}^{-1}. Determine the magnetic force acting on the α\alpha-particle.

Options

  1. A9.6×10−12 N9.6 \times 10^{-12} \text{ N} towards west
  2. B9.6×10−12 N9.6 \times 10^{-12} \text{ N} towards east
  3. C4.8×10−12 N4.8 \times 10^{-12} \text{ N} towards east
  4. D4.8×10−12 N4.8 \times 10^{-12} \text{ N} towards west

Correct answer

A. 9.6×10−12 N9.6 \times 10^{-12} \text{ N} towards west

Step-by-step solution

Charge of an alpha particle, q=+2e=3.2×10−19 Cq = +2e = 3.2 \times 10^{-19} \text{ C}

Velocity, v=3.0×104 km s−1=3.0×107 m s−1v = 3.0 \times 10^4 \text{ km s}^{-1} = 3.0 \times 10^7 \text{ m s}^{-1}

Magnetic field, B=1.0 TB = 1.0 \text{ T}

Let East be i^\hat{i}, North be j^\hat{j}, and vertically upward be k^\hat{k}.

v⃗=3.0×107k^ m s−1\vec{v} = 3.0 \times 10^7 \hat{k} \text{ m s}^{-1}

B⃗=1.0j^ T\vec{B} = 1.0 \hat{j} \text{ T}

Magnetic force, F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v} \times \vec{B})

F⃗=3.2×10−19(3.0×107k^×1.0j^)\vec{F} = 3.2 \times 10^{-19} (3.0 \times 10^7 \hat{k} \times 1.0 \hat{j})

F⃗=9.6×10−12(k^×j^) N\vec{F} = 9.6 \times 10^{-12} (\hat{k} \times \hat{j}) \text{ N}

Since k^×j^=−i^\hat{k} \times \hat{j} = -\hat{i}, the force is directed towards the west.

Magnitude of force is 9.6×10−12 N9.6 \times 10^{-12} \text{ N}.
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