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Physics · 2025 · 3 marks
Moving Charges and MagnetismMagnetic Field due to a Current Element, Biot-Savart Law3 marksApplyshort answer
CBSE 2025 · Region 1 · Set 3 · Q23
A particle of charge q is moving with a velocity v at a distance ' d ' from a long straight wire carrying a current ' I ' as shown in figure. At this…
A particle of charge $\displaystyle q$ is moving with a velocity $\displaystyle \vec{v}$ at a distance ' $\displaystyle d$ ' from a long straight wire carrying a current ' $\displaystyle I$ ' as shown in figure. At this instant, it is subjected to a uniform electric field $\displaystyle \overrightarrow{\mathrm{E}}$ such that the particle keeps moving undeviated. In terms of unit vectors $\displaystyle \hat{i}, \hat{j}$ and $\displaystyle \hat{k}$, find -
(a)
the magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$,
(b)
the magnetic force $\displaystyle \overrightarrow{\mathrm{F}}_{\mathrm{m}}$, and
(c)
the electric field $\displaystyle \overrightarrow{\mathrm{E}}$, acting on the charge.
Marking-scheme solution
(a)
$\displaystyle \vec{B}=\frac{\mu_{0} I}{2 \pi d}(-\hat{k})$
(b)
$\displaystyle \vec{F}_{B}=q \vec{v} \times \vec{B}=\frac{q v \mu_{0} I}{2 \pi d}(-\hat{j})$
(c)
$\displaystyle \vec{F}_{e}=-\vec{F}_{B}$ (for undeviated motion of the charged particle)
$\displaystyle \therefore \vec{F}_{e}=\frac{q v \mu_{0} I}{2 \pi d}(\hat{j})$
$\displaystyle \vec{F}_{e}=q \vec{E}$
$\displaystyle \therefore \vec{E}=\frac{\mu_{0} v I}{2 \pi d} \hat{j}$
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