Chapter 18 · NEET Preparation Physics

Moving Charges and Magnetism

25 practice questions · 5 marks weightage

Topics in this chapter

  • Magnetic Fields From Currents
  • Magnetic Forces and Charged Motion
  • Forces, Torque, and Magnetic Dipoles
  • Galvanometers and Magnetic Measurements

Cheat sheet

Cheat Sheet

Practice questions

  1. Question 1Q-E44E5A

    A proton and an alpha particle are projected perpendicular to a uniform magnetic field with the same kinetic energy. The ratio of the radii of their circular paths (rp:rαr_p : r_\alpha) is:

    1. A.1 : 1
    2. B.1 : \sqrt{2}
    3. C.\sqrt{2} : 1
    4. D.2 : 1
    See the answer and solution
  2. Question 2Q-53B5D1

    A circular coil of radius RR and NN turns carries a current II. The magnetic field at its center is BB. At what distance from the center along the axis of the coil will the magnetic field be B/8B/8?

    1. A.3R\sqrt{3}R
    2. B.2R\sqrt{2}R
    3. C.2R2R
    4. D.3R3R
    See the answer and solution
  3. Question 3Q-173A54

    A galvanometer of resistance GG is converted into a voltmeter to measure up to VV volts by connecting a resistance R1R_1 in series. If it is to be converted to measure up to 2V2V volts, the series resistance required is R2R_2. The relation between R1R_1 and R2R_2 is:

    1. A.R2=2R1+GR_2 = 2R_1 + G
    2. B.R2=2R1−GR_2 = 2R_1 - G
    3. C.R2=R1+GR_2 = R_1 + G
    4. D.R2=2R1R_2 = 2R_1
    See the answer and solution
  4. Question 4Q-95DBA0

    A long straight wire carries a current II. The magnetic field at a perpendicular distance rr from the wire is BB. If the current is doubled and the distance is halved, the new magnetic field becomes:

    1. A.B/4B / 4
    2. B.B/2B / 2
    3. C.2B2B
    4. D.4B4B
    See the answer and solution
  5. Question 5Q-ED1BE1

    A rectangular loop of sides 10 cm10\text{ cm} and 5 cm5\text{ cm} carrying a current 2 A2\text{ A} is placed in a uniform magnetic field of magnitude 0.3 T0.3\text{ T}. The maximum torque acting on the loop is:

    1. A.3×10−3 N m3 \times 10^{-3}\text{ N m}
    2. B.6×10−3 N m6 \times 10^{-3}\text{ N m}
    3. C.3×10−2 N m3 \times 10^{-2}\text{ N m}
    4. D.1.5×10−2 N m1.5 \times 10^{-2}\text{ N m}
    See the answer and solution

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