JEE Advanced 2017 · Paper 2 · Physics Q8

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PhysicsMultiple correct+4 / -2Question 8 of 54

A uniform magnetic field $B$ exists in the region between $x = 0$ and $x = \frac{3R}{2}$ (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge $+Q$ and momentum $p$ directed along $x$ -axis enters region 2 from region 1 at point $P_1$ $(y = -R)$ . Which of the following option(s) is/are correct?

Region 1 (x < 0), region 2 (0 < x < 3R/2, uniform field B into the page, crosses) and region 3 (x > 3R/2). A charge +Q moving along +x enters region 2 at P₁ on the y-axis at y = −R. P₂ is where the x-axis meets the boundary x = 3R/2.

For

B > \frac{2}{3}\frac{p}{QR}

, the particle will re-enter region 1

For

B = \frac{8}{13}\frac{p}{QR}

, the particle will enter region 3 through the point

P_2

x

-axis

When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point

P_1

and the farthest point from

y

-axis is

p/\sqrt{2}

For a fixed

B

, particles of same charge

Q

and same velocity

v

, the distance between the point

P_1

and the point of re-entry into region 1 is inversely proportional to the mass of the particle

Answer and worked solution

Official answer: A, B

Radius. $r=\dfrac p{QB}$ . At $P_1$ the force $Q\vec v\times\vec B$ points along $+y$ , so the particle curves upward. The centre of its circle is at $(0,-R+r)$ .

(A) It turns back into region 1 if the circle never reaches $x=\tfrac{3R}2$ , i.e. $r<\tfrac{3R}2$ , i.e. $B>\dfrac2{3}\dfrac p{QR}$ . True.
(B) With $B=\dfrac8{13}\dfrac p{QR}$ , $r=\tfrac{13R}8$ and the centre is at $(0,\tfrac{5R}8)$ . At $x=\tfrac{3R}2$ :
$\left(y-\tfrac{5R}8\right)^2=\tfrac{169-144}{64}R^2\implies y=0\ \text{(first crossing)}.$
It exits at $P_2$ . True.
(C) On the longest such path ( $r=\tfrac{3R}2$ ), at the point farthest from the $y$ -axis the momentum has turned from $+x$ to $+y$ . $|\Delta\vec p|=\sqrt2\,p$ , not $p/\sqrt2$ . False.
(D) The re-entry distance is $2r=\dfrac{2mv}{QB}\propto m$ , directly, not inversely. False.

(A), (B)

Topics:Motion in magnetic and electric fields Force on a moving charge

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JEE Advanced 2017 · Paper 2, Physics question 8