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MOVING CHARGES AND MAGNETISM - 8

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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.  The field due to a wire of n turns and radius r which carries a current I is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction




  •  3/2 h^2/r^2
  •  2/3 h^2/r^2
  •  3/2 r^2/h^2
  •  2/3 r^2/h^2
Solution









Q2.A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction being parallel to the axis along east to west. A current carrying wire in north-south direction passes through this region. The wire intersects the axis and experiences a force of 1.2 N downward. If the wire is turned from north south to northeast-southwest direction, then magnitude and direction of force is





  •  1.2 N, upward
  •  1.2√2, downward
  •  1.2 N, downward
  •  1.2/√2 N, downward
Solution










Q3.  A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the centre. The value of the magnetic induction at the centre due to current in the ring is



  •  Proportional to 2(180°-θ)
  •  Inversely proportional to r
  •  Zero, only if θ=180°
  •  Zero for all values of θ
Solution











Q4. A moving coil galvanometer is based on the
  •  heating effect of current
  •  magnetic effect of current
  •  chemical effect of current
  •  Peltier effect of current
Solution












Q5. A parallel plate capacitor is moving with a velocity of 25 ms^(-1) through a uniform magnetic field of 4.0 T as shown in figure. If the electric filed within the capacitor plates is 175 NC^(-1) and plate area is 2.5×10^(-7) m^2, then the magnetic force experienced by the positive charge plate is    

 




  •  8.85×10^(-13) N directed out of the plane of the paper
  •  zero
  •  8.85×10^(-15) N directed out of the plane of the paper
  •  none of these
Solution
 










Q6. A metallic block carrying current I is subjected to a uniform magnetic induction B as shown in the figure. The moving charges experience a force F given by …… which results in the lowering of the potential of the face ….. Assume the speed of the carriers to be v





  •  eVBk ̂,ABCD
  •  eVBk ̂,EFGH
  • -eVBk ̂,ABCD
  •  -eVBk ̂,EFGH
Solution












Q7.From a cylinder of radiusR, a cylinder of radius R/2 is removed, as shown in figure. Current flowing in the remaining cylinder is I. Then, magnetic field strength is




  •  Zero at point A
  •  Zero at point B
  •  (μ0 T)/2πR at point A
  •  (μ0 I)/3πR at point B
Solution













Q8.A uniform conducting rectangle loop of sides l,b and mass m carrying current i is hanging horizontally with the help of two vertical strings. There exists a uniform horizontal magnetic filed B which is parallel to the longer side of loop. The value of tension which is least is


  •  (mg-Bb)/2
  •  T(mg+Bb )/2
  •  (mg-2iBb)/2
  •  (mg+2Bb)/2
Solution













Q9.A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ=3x,PR=4 x and QR=5x. If the magnitude of the magnetic field at P due to this loop is k((μ0 I)/(48π x)), find the value of k
  •  8
  •  3
  •  7
  •  none
Solution












Q10. Two very long, straight, parallel wires carry steady currents I and –I, respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires. Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is

  •  (μ0 Iqv)/2πd
  •  (μ0 Iqv)/πd
  •  (2μ0 Iqv)/πd
  • 0
Solution









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