A charge of 4.0 is moving with a velocity of 4.0 × 10⁶ ms⁻¹ along the positive y axis under a magnetic field B of strength (2k) T. The force acting on the charge is xiN. The value of x is __________. Numerical answer type. Work it out, then open the answer.

Answer: 32. F = q(v × B) = 4 × 10⁻⁶ × (4 × 10⁶ j × 2k) j × k = i, so F = 4 × 10⁻⁶ × 8 × 10⁶ i = 32i N x = 32

A charge of 4.0 is moving with a velocity of 4.0 × 10⁶ ms⁻¹ along the… | JEE Main 2024 (29 Jan, Shift 2)

Q30JEE Main 2024 · 29 JAN · SHIFT 2Easy

Moving Charge in a Magnetic Field

A charge of $4.0\ \mu\text{C}$ is moving with a velocity of $4.0 \times 10^{6}\text{ ms}^{-1}$ along the positive $y$ axis under a magnetic field $\vec{B}$ of strength $(2\hat{k})\text{ T}$ . The force acting on the charge is $x\hat{i}N$ . The value of $x$ is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 32

$\vec{F} = q(\vec{v} \times \vec{B}) = 4 \times 10^{-6} \times (4 \times 10^6 \hat{j} \times 2\hat{k})$
$\hat{j} \times \hat{k} = \hat{i}$ , so $\vec{F} = 4 \times 10^{-6} \times 8 \times 10^6 \hat{i} = 32\hat{i}$ N
$x = 32$

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