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4.Moving Charges and Magnetism
normal
In the $xy$-plane, the region $y>0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $\mathrm{y}<0$ has another uniform magnetic field $\mathrm{B}_2 \hat{k}$. A positively charged particle is projected from the origin along the positive $y$-axis with speed $v_0=\pi \mathrm{m} \mathrm{s}^{-1}$ at $\mathrm{t}=0$, as shown in the figure. Neglect gravity in this problem. Let $\mathrm{t}=\mathrm{T}$ be the time when the particle crosses the $\mathrm{x}$-axis from below for the first time. If $B_2=4 B_1$, the average speed of the particle, in $\mathrm{m} \mathrm{s}^{-1}$, along the $\mathrm{x}$-axis in the time interval $\mathrm{T}$ is. . . .
(IMAGE)
A$1$
B$0$
C$2$
D$5$
(IIT-2018)
Solution
$\mathrm{R}_1=\frac{\mathrm{mv}}{\mathrm{qB}}$(image)
$\mathrm{R}_2=\frac{\mathrm{mv}}{\mathrm{qB}_2}$
$\mathrm{~B}_2=4 \mathrm{~B}_1$
$\mathrm{R}_2=\frac{1}{4} \mathrm{R}_1$
Distance traveled in $x$ direction
$\Delta \mathrm{x}=2 \mathrm{R}_1+2 \mathrm{R}_2$
$\Delta \mathrm{x}=2 \mathrm{R}_1+\frac{\mathrm{R}_1}{2}=\frac{5 \mathrm{R}_1}{2}$
$\mathrm{~T}_1=\frac{\pi \mathrm{m}}{\mathrm{qB}_1}$
$\mathrm{~T}_2=\frac{\pi \mathrm{m}}{\mathrm{qB}_2}=\frac{\mathrm{T}_1}{4}$
Total time $=\frac{\mathrm{T}_1}{2}+\frac{\mathrm{T}_2}{2}=\frac{5 \mathrm{~T}_1}{8}$
Average speed $\mathrm{V}=\frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}=\frac{\left(\frac{5 \mathrm{R}_1}{2}\right)}{\left(\frac{5 \mathrm{~T}_1}{8}\right)}=4 \frac{\mathrm{R}_1}{\mathrm{~T}_1}$
$\mathrm{R}=\frac{\mathrm{mV}}{\mathrm{qB}}$
$\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{qB}}$
Average speed $=\frac{4 \mathrm{R}}{\mathrm{T}}=\frac{4 \mathrm{~V}}{2 \pi}=2$
Average speed $=2$
Standard 12
Physics
