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MOVING CHARGES AND MAGNETISM

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JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.

Q1. Two particles A and B of masses mA and mB, respectively, and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB, respectively and the trajectories are as shown in figure. Then


  •   mAvA < mBvB
  •  mA vA>mB vB
  •  mA<mb
  •  mA=mB and vA=vB
Solution


Q2.An electron is accelerated from rest through a potential difference V. This electron experiences a force F in a uniform magnetic field. On increasing the potential difference to V', the force experienced by the electron in the same magnetic field becomes 2F. Then, the ratio (V^'/V) is equal to
  •  1/4
  •  2/1
  •  1/2
  •  4/1
Solution



Q3.  A charged particle moving along +ve x-direction with a velocity v enters a region where there is a uniform magnetic field B_0 (-k ̂ ), from x=0 to x=d. The particle gets deflected at an angle θ from its initial path. The specific charge of the particle is


  •   (v cos⁡θ)/Bd
  •  (v tan⁡θ)/Bd
  •  v/Bd
  •  (v sin⁡θ)/Bd
Solution



Q4. A long cylindrical wire of radius ‘a’ carries a current i distributed uniformly over its cross section. If the magnetic fields at distances r<a and R<a from the axis have equal magnitude, then
  •  a=(R+r)/2
  •  a=√Rr
  •  a=Rr/R+r
  •  a=R^2/r
Solution




Q5. A Q5.charged particle moves with velocity v =ai ̂+dj ̂ in a magnetic field B =Ai ̂+Dj ̂. The force acting on the particle has magnitude F. Then,

 

  •  F=0, if aD=dA
  •  F=0, if aD=-dA
  •  F=, if aA=-dD
  •  F∝(a^2+b^2 )^(1/2)×(A^2+D^2 )^(1/2)
Solution
 



Q6. A very thin metallic wires placed along X-and Y-axes carry equal currents as shown in figure AB and CD are lines at 45° with the axes. The magnetic fields will be zero on the line


  •  AB
  •  CD
  • Segment OB only of line AB
  •  Segment OC only of line CD
Solution




Q7.A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the X-Y plane and a steady current I flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

  •  (μ_0 NI)/(2(b-a)) ln⁡(b/a)
  •  (μ_0 NI)/(2(b-a)) ln⁡((b+a)/(b-a))
  •  (μ_0 NI)/2b ln⁡(b/a)
  •  (μ_0 NI)/2b ln⁡((b+a)/(b-a))
Solution




Q8.An electron is projected at an angle θ with a uniform magnetic field. If the pitch of the helical path is equal to its radius, then the angle of projection is
  •  tan^(-1)⁡Ï€
  •  Ttan^(-1)⁡2Ï€
  •  cot^(-1)⁡Ï€
  •  cot^(-1)⁡2Ï€
Solution




Q9.An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR. The magnetic field due to this current at the point M is H_1. Now, another infinitely long straight conductor QS is connected to Q so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now H_2. The ratio H_1/H_2 is given by

  •  1/2
  •  1
  •  2/3
  •  2
Solution





Q10. A small block of mass m, having charge q, is placed on a frictionless inclined plane making an angle θ with the horizontal. There exists a uniform magnetic field B parallel to the inclined plane but perpendicular to the length of spring. If m is slightly pulled on the incline in downward direction, the time period of oscillation will be (assume that the block does not leave contact with the plane)


  •  2Ï€√(m/K)
  •  2Ï€√(2m/K)
  •  2Ï€√(qB/K)
  • 2Ï€√(qB/2K)
Solution


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