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Magnetism question

2013 · Shift 1 · Q58
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Magnetism question

2013 · Shift 1 · Q58

JEE AdvancedPhysicsMagnetismMultiple correct+4 / −2
A particle of mass M and positive charge Q, moving with a constant velocity u→1=4i^{\overrightarrow u _1} = 4\widehat iu1​=4i ms −-− 1 enters a region of uniform static magnetic field, normal to the xy plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 ms with a velocity u→2=2(3i^+j^){\overrightarrow u _2} = 2\left( {\sqrt 3 \widehat i + \widehat j} \right)u2​=2(3​i+j​) ms −-− 1. The correct statement(s) is(are)
  1. A
    The direction of the magnetic field is −-− z direction.
  2. B
    The direction of the magnetic field is +z direction.
  3. C
    The magnitude of the magnetic field is 50πM3Q{{50\pi M} \over {3Q}}3Q50πM​ units.
  4. D
    The magnitude of the magnetic field is 100πM3Q{{100\pi M} \over {3Q}}3Q100πM​ units.
View written solutionFree

Correct answer: A, C

  1. Given data
  • Initial velocity: u⃗1=4i^ m s−1\vec u_1 = 4\hat i\ \text{m s}^{-1}u1​=4i^ m s−1
  • Final velocity after emerging: u⃗2=2(3i^+j^)=23i^+2j^ m s−1\vec u_2 = 2(\sqrt3\hat i+\hat j)=2\sqrt3\hat i+2\hat j\ \text{m s}^{-1}u2​=2(3​i^+j^​)=23​i^+2j^​ m s−1
  • Time spent in magnetic field: t=10 ms=10−2 st=10\text{ ms}=10^{-2}\text{ s}t=10 ms=10−2 s
  • Magnetic field is uniform and perpendicular to the xyxyxy-plane.
  1. Use the fact that magnetic force does not change speed

In a magnetic field, only direction of velocity changes, not its magnitude.

Initial speed: ∣u⃗1∣=4|\vec u_1|=4∣u1​∣=4

Final speed: ∣u⃗2∣=(23)2+22=12+4=4|\vec u_2|=\sqrt{(2\sqrt3)^2+2^2}=\sqrt{12+4}=4∣u2​∣=(23​)2+22​=12+4​=4

So this is consistent with motion in a uniform magnetic field.

  1. Find the angle through which velocity turns

Initial velocity is along +x+x+x.

For final velocity, tan⁡θ=223=13\tan\theta=\frac{2}{2\sqrt3}=\frac1{\sqrt3}tanθ=23​2​=3​1​ So, θ=30∘=π6\theta=30^\circ=\frac{\pi}{6}θ=30∘=6π​

Thus the velocity direction changes from 0∘0^\circ0∘ to 30∘30^\circ30∘.

  1. Determine the direction of magnetic field

Initially, velocity is along +x+x+x. Magnetic force is F⃗=Q(v⃗×B⃗)\vec F=Q(\vec v\times \vec B)F=Q(v×B) with Q>0Q>0Q>0.

Since the particle emerges with a positive yyy-component of velocity, the trajectory bends toward +y+y+y. So initially the magnetic force must be toward +y+y+y.

Now check both possibilities:

  • If B⃗=+Bk^,\vec B=+B\hat k,B=+Bk^, then i^×k^=−j^\hat i\times \hat k=-\hat ji^×k^=−j^​ so force would be toward −y-y−y. This is not correct.

  • If B⃗=−Bk^,\vec B=-B\hat k,B=−Bk^, then i^×(−k^)=+j^\hat i\times(-\hat k)=+\hat ji^×(−k^)=+j^​ so force is toward +y+y+y. This is correct.

Therefore, the magnetic field is in the −z-z−z direction.

So, A is correct and B is incorrect.

  1. Find magnitude of magnetic field

In uniform magnetic field, angular speed is ω=QBM\omega=\frac{QB}{M}ω=MQB​

The velocity vector rotates uniformly. In time ttt, the angle turned is θ=ωt=QBMt\theta=\omega t=\frac{QB}{M}tθ=ωt=MQB​t

Here, θ=π6,t=10−2 s\theta=\frac{\pi}{6},\qquad t=10^{-2}\text{ s}θ=6π​,t=10−2 s

So, QBM(10−2)=π6\frac{QB}{M}(10^{-2})=\frac{\pi}{6}MQB​(10−2)=6π​

Hence, B=MQ⋅π6×10−2B=\frac{M}{Q}\cdot\frac{\pi}{6\times10^{-2}}B=QM​⋅6×10−2π​

B=MQ⋅100π6B=\frac{M}{Q}\cdot\frac{100\pi}{6}B=QM​⋅6100π​

B=50πM3QB=\frac{50\pi M}{3Q}B=3Q50πM​

Therefore, C is correct and D is incorrect.

  1. Final option check
  • A: Correct
  • B: Incorrect
  • C: Correct
  • D: Incorrect

Hence the correct statements are: A, C\boxed{A,\ C}A, C​

  1. Comparison with stored answer

Stored correct answer: A, C

This matches the derived answer exactly.

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