JEE Advanced 2018 · Paper 1 · Physics Q12

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PhysicsNumerical+3Question 12 of 54

In the $xy$ -plane, the region $y > 0$ has a uniform magnetic field $B_1\hat k$ and the region $y < 0$ has another uniform magnetic field $B_2\hat k$ . A positively charged particle is projected from the origin along the positive $y$ -axis with speed $v_0 = \pi\ \mathrm{m\,s^{-1}}$ at $t = 0$ , as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$ -axis from below for the first time. If $B_2 = 4B_1$ , the average speed of the particle, in $\mathrm{m\,s^{-1}}$ , along the $x$ -axis in the time interval $T$ is ______.

The x-y plane: field B₁ (lighter shading) for y > 0 and B₂ (darker) for y < 0. A particle starts at the origin moving along +y with speed v₀ = π m s⁻¹.

Your answer

Answer and worked solution

Official answer: 1.2

Path. At the origin $\vec v=v_0\hat j$ and $\vec F=q\vec v\times B_1\hat k\propto\hat i$ . The particle makes a semicircle of radius $r_1=\dfrac{mv_0}{qB_1}$ in $y>0$ and returns to the $x$ -axis at $x=2r_1$ .

Below the axis it makes a semicircle the other way, of radius $r_2=\dfrac{mv_0}{qB_2}=\dfrac{r_1}4$ , and crosses back upwards at $x=2r_1-2r_2$ .

Time.
$T=\frac{\pi r_1}{v_0}+\frac{\pi r_2}{v_0}=\frac{\pi(r_1+r_2)}{v_0}.$

Net $x$ -displacement over time (the reading of the official answer 1.2):
$\frac{2(r_1-r_2)}T=\frac{2v_0}\pi\cdot\frac{r_1-r_2}{r_1+r_2}=\frac{2\pi}\pi\cdot\frac{3/4}{5/4}=1.2\text{ m/s}.$

Total $x$ -distance over time (the other accepted reading):
$\frac{2(r_1+r_2)}T=\frac{2v_0}\pi=2\text{ m/s}.$

Answer: 1.2 or 2 (both accepted)

Topics:Motion in magnetic and electric fields Uniform circular motion

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