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Magnetics question

2019 · 11 Jan · Shift 2 · Q69
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Magnetics question

2019 · 11 Jan · Shift 2 · Q69

JEE MainPhysicsMagneticsMCQ+4 / −1
The region between y = 0 and y = d contains a magnetic field B→=Bz^\overrightarrow B = B\widehat zB=Bz. A particle of mass m and charge q enters the region with a velocity v→=vi^.\overrightarrow v = v\widehat i.v=vi. If d =mv2qB,={{mv} \over {2qB}},=2qBmv​, the acceleration of the charged particle at the point of its emergence at the other side is :
  1. A
    qvBm(−32i^−12j^){{qvB} \over m}\left( -{{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)mqvB​(−23​​i−21​j​)
  2. B
    qvBm(12i^−32j^){{qvB} \over m}\left( {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right)mqvB​(21​i−23​​j​)
  3. C
    qvBm(−j^+i^2){{qvB} \over m}\left( {{{ - \widehat j + \widehat i} \over {\sqrt 2 }}} \right)mqvB​(2​−j​+i​)
  4. D
    qvBm(j^+i^2){{qvB} \over m}\left( {{{\widehat j + \widehat i} \over {\sqrt 2 }}} \right)mqvB​(2​j​+i​)
View written solutionFree

Correct answer: A

  1. Magnetic force and motion inside the field

A charged particle enters a region where B⃗=Bz^\vec B = B\hat zB=Bz^ and initial velocity is v⃗=vi^.\vec v = v\hat i.v=vi^.

The magnetic force is F⃗=qv⃗×B⃗.\vec F = q\vec v \times \vec B.F=qv×B. Initially, v⃗×B⃗=vi^×Bz^=vB(i^×z^)=−vBj^.\vec v \times \vec B = v\hat i \times B\hat z = vB(\hat i \times \hat z)= -vB\hat j.v×B=vi^×Bz^=vB(i^×z^)=−vBj^​. So the particle bends toward negative yyy for positive qqq. Since the problem asks for emergence at the other side y=dy=dy=d, the relevant geometry is the circular motion in the field region, and the acceleration direction at emergence is given by the Lorentz force.

The speed remains constant in a magnetic field, and the particle moves along a circular arc of radius R=mvqB.R=\frac{mv}{qB}.R=qBmv​.

Given d=mv2qB=R2.d=\frac{mv}{2qB}=\frac R2.d=2qBmv​=2R​.

  1. Find the angle turned before emerging

Let the particle enter at y=0y=0y=0. Inside the field it follows a circle of radius RRR.

Take the center of the circle such that the vertical rise of the particle from entry point to exit point is ddd. For circular motion starting with tangent along +x+x+x, the change in height after turning through angle θ\thetaθ is y=R(1−cos⁡θ).y=R(1-\cos\theta).y=R(1−cosθ).

At emergence, R(1−cos⁡θ)=d=R2.R(1-\cos\theta)=d=\frac R2.R(1−cosθ)=d=2R​. Hence 1−cos⁡θ=121-\cos\theta=\frac121−cosθ=21​ cos⁡θ=12\cos\theta=\frac12cosθ=21​ θ=π3.\theta=\frac\pi3.θ=3π​.

So the velocity direction has rotated by 60∘60^\circ60∘ clockwise from the initial +x+x+x direction. Therefore,

= v\left(\frac12\hat i-\frac{\sqrt3}{2}\hat j\right).$$ 3. **Acceleration at emergence** Magnetic acceleration is $$\vec a=\frac{q}{m}(\vec v\times \vec B).$$ Now at exit, $$\vec v = v\left(\frac12\hat i-\frac{\sqrt3}{2}\hat j\right), \qquad \vec B=B\hat z.$$ Compute cross product: $$\vec v\times \vec B = vB\left(\frac12\hat i-\frac{\sqrt3}{2}\hat j\right)\times \hat z.$$ Using $$\hat i\times \hat z=-\hat j, \qquad \hat j\times \hat z=\hat i,$$ we get $$\vec v\times \vec B = vB\left[\frac12(-\hat j)-\frac{\sqrt3}{2}(\hat i)\right].$$ So $$\vec a=\frac{qvB}{m}\left(-\frac{\sqrt3}{2}\hat i-\frac12\hat j\right).$$ 4. **Match with options** This is exactly **Option A**: $$\boxed{\frac{qvB}{m}\left(-\frac{\sqrt3}{2}\hat i-\frac12\hat j\right)}.$$
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