Magnetic Field Due to a Current-Carrying Conductor (Right-Hand Thumb Rule) MCQs
Practice Magnetic Field Due to a Current-Carrying Conductor (Right-Hand Thumb Rule) multiple-choice questions from Magnetic Effects of Electric Current (Class 10 Science) - tap an answer for instant feedback and a step-by-step solution. Practice the full set free on the RankByte app.
Magnetic Field Due to a Current-Carrying Conductor (Right-Hand Thumb Rule)Quiz - Solve & Score
Q1. Consider the following statements about the field of a straight conductor: (I) The field forms concentric circles. (II) Its direction is given by the right-hand thumb rule. (III) Reversing the current has no effect on the field. Which of the above statement(s) is/are correct?
- A.Only III
- B.Only I and II
- C.None of these
- D.Only I
Answer: B. Only I and II
What does the chapter say about this? Correct: I, II. Does that line up with one of the options? Yes - Correct. And the rest? option A) 'Only III' fails since Re-check each statement individually; option C) 'None of these' misses the point - Re-check each statement individually. Pick: B) Only I and II.
Q2. An overhead transmission line carries a steady 11 A; what magnetic field does it create at a spot 40 cm from it?
- A.3.46*10^-5 T
- B.5.5*10^-6 T
- C.8.8*10^-7 T
- D.1.37*10^-5 T
Answer: B. 5.5*10^-6 T
For a long straight wire B = mu0*I/(2*pi*r) with mu0/(2*pi) = 2*10^-7 T-m/A. B = 2*10^-7 * 11 / 0.4 = 5.5*10^-6 T.
Q3. A welder draws 20 A through a long straight lead; how strong is the field at a point 3 cm to one side of the wire?
- A.8.38*10^-4 T
- B.1.33*10^-4 T
- C.4.44*10^-3 T
- D.1.2*10^-7 T
Answer: B. 1.33*10^-4 T
For a long straight wire B = mu0*I/(2*pi*r) with mu0/(2*pi) = 2*10^-7 T-m/A. B = 2*10^-7 * 20 / 0.03 = 1.33*10^-4 T.
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