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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Advanced 2013 · Paper 1 · Physics · Magnetic Effects of Current

A particle of mass M and positive charge Q, moving with a constant...

A particle of mass M and positive charge Q, moving with a constant velocity \overset{\to}{\text{u}}1 = 4\overset{^}{\text{i}}{\text{ms}}^{- 1}, enters a region of uniform static magnetic field normal to the x-y 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 milliseconds with a velocity \overset{\to}{\text{u}}2 = 2\left(\sqrt{3}\overset{^}{\text{i}} + \overset{^}{\text{j}}\right){\text{ms}}^{- 1}. The correct statement (s) is (are)A particle of mass M and positive charge Q, moving with a constant velocity → u 1 = 4 ^ i ms ^ - 1 , enters a region of

Options

  1. AThe direction of the magnetic field is -z direction.
  2. BThe direction of the magnetic field is +z direction.
  3. CThe magnitude of the magnetic field \dfrac{5 0 π \text{M}}{3 \text{Q}}units.
  4. DThe magnitude of the magnetic field \dfrac{10 0 π \text{M}}{3 \text{Q}}units.

Correct answer

A. The direction of the magnetic field is -z direction.

Step-by-step solution

A particle of mass M and positive charge Q, moving with a constant velocity → u 1 = 4 ^ i ms ^ - 1 , enters a region of

the angular displacement(θ),\left(\theta\right),
θ=ωt→(1)</p><p>\theta = \omega \text{t}\to \left(1\right)</p><p>\omega =angularvelocity,angular velocity,t =timetakenbythetheparticleemergesontheotherside</p><p>time taken by the the particle emerges on the other side </p><p>t = 10ms = 10 \times {10}^{- 3}S

angular velocity of particleinamagneticfield(ω),particle in a magnetic field \left(\omega\right),\omega = qBm\dfrac{qB}{m}\to \left(2\right)</p><p></p><p>q =charge,charge,B =magneticfield,magnetic field,m =massoftheparticle.<br/>substituteequation2inequation1<br/>mass of the particle.<br />substitute equation 2 in equation 1<br />\theta = \dfrac{\text{qB}}{\text{m}} \text{t}<br/>Anglebetweenvelocityvectorsis(θ)=π6<br />Angle between velocity vectors is \left(\theta\right) = \dfrac{\pi}{6}
Solving above we get B=50πm3q\text{B} = \dfrac{5 0 \pi \text{m}}{3 \text{q}}

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