Magnetic Fields Produced by Currents: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A thin wire of length L carries a steady current I. In case 1 it is bent into a square loop; in case 2 the same wire is bent into a single circular loop. Each loop is closed (the leads are twisted together and contribute no field). Let Bsq and Bc be the magnitudes of the magnetic field at the centres of the square and the circle respectively. The ratio Bsq/Bc is ______. Exactly one option is correct.
A82π2
Bπ242
Cπ282
Dπ22
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Answer: C
Square of side a=L/4: each side is at perpendicular distance a/2 from the centre and subtends 45∘ on either side of the perpendicular, so by Biot–Savart each side gives 4π(a/2)μ0I(sin45∘+sin45∘)=2πa2μ0I. By the right-hand rule (curl the fingers along the current) all four sides give the same direction, so Bsq=πa22μ0I=πL82μ0I.
Circle of radius R=L/(2π): Bc=2Rμ0I=Lπμ0I.
Ratio =π282≈1.146. The square gives the larger field because parts of its wire lie closer to the centre (checked by numerical Biot–Savart integration).
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
Use right-handed axes x,y,z; the +z axis points towards a viewer looking down on the xy-plane. A long thin wire carries current I along the following path, all in the xy-plane: it comes in along the line x=+R from y=+∞ down to the point (R,0); it then follows the semicircle of radius R centred at the origin O through the point (0,−R) to the point (−R,0); finally it goes out along the line x=−R from (−R,0) to y=+∞. The magnetic field at O is ______. Exactly one option is correct.
A4πRμ0I(π+2) along −z
B4πRμ0I(π+2) along +z
C4πRμ0I(π−2) along −z
D4πRμ0I(π+4) along −z
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Answer: A
Semicircle: Biot–Savart for an arc of angle π gives 4Rμ0I. Going from (R,0) through (0,−R) to (−R,0) is clockwise as seen from +z; curling the right-hand fingers clockwise points the thumb along −z.
Each straight wire: O lies at perpendicular distance R from the wire, on the perpendicular through its finite end, so B=4πRμ0I(sin90∘−sin0∘)=4πRμ0I.
Directions: wire at x=+R carries current along −y^, and O is on its −x^ side: (−y^)×(−x^)=−z^. Wire at x=−R carries +y^, O on its +x^ side: y^×x^=−z^. All three add:
B=4Rμ0I+4πR2μ0I=4πRμ0I(π+2) along −z.
Trap: treating the two antiparallel leads as cancelling gives (π−2); they lie on opposite sides of O, so their fields add. (Verified by numerical integration.)
Difficulty: Very hard
Question 3Level 2 · ExamSingle correct
Two flat circular coils carry the same current I and have the same number of turns. Coil 1 has radius R; coil 2 has radius 2R. At what distance x from the centre of coil 1, measured along its axis, does the magnitude of the field of coil 1 equal the magnitude of the field of coil 2 at its own centre? Exactly one option is correct.
AR3
BR42/3−1
CR/2
DR22/3−1
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Answer: D
Axis of a coil (N turns): B(x)=2(R2+x2)3/2μ0NIR2. Coil 2 at its centre: 2(2R)μ0NI=4Rμ0NI.
Setting equal: 2(R2+x2)3/2R2=4R1⇒(R2+x2)3/2=2R3⇒R2+x2=22/3R2,
so x=R22/3−1≈0.766R.
Check: the axial field falls monotonically from μ0NI/(2R) at x=0 to 0, so the target value μ0NI/(4R) (half the central value) is reached exactly once for x>0.
Trap: writing (R2+x2)3/2=4R3 (forgetting the factor 2 in the denominator) gives R42/3−1.
Difficulty: Hard
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