Biot-Savart Law & Field due to Current — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Biot-Savart Law & Field due to Current MCQs with step-by-step solutions (21 questions). Part of Magnetic Fields due to Electric Current. Practise online on Prepizo — no login needed.
▶ Practise Biot-Savart Law & Field due to Current online (free)
Questions with solutions
Q1 — Biot-Savart Law & Field due to Current · medium · theory
According to the Biot-Savart law, the field due to a current element $I\,dl$ at distance $r$ is:
A. $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\cos\theta}{r^2}$
B. $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$ ✓ Correct
C. $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r}$
D. $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{r^3}$
Solution: The contribution is largest perpendicular to the element and vanishes along its own direction.
Q2 — Biot-Savart Law & Field due to Current · easy · theory
The magnetic field at a perpendicular distance $r$ from a long straight conductor carrying current $I$ is:
A. $\dfrac{\mu_0I}{r^2}$
B. $\dfrac{\mu_0I}{4\pi r}$
C. $\dfrac{\mu_0I}{2\pi r}$ ✓ Correct
D. $\dfrac{\mu_0I}{2r}$
Solution: The field circles the wire, falling off inversely with distance.
Q3 — Biot-Savart Law & Field due to Current · easy · theory
The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is:
A. $\dfrac{\mu_0I}{4R}$
B. $\dfrac{\mu_0I}{2R}$ ✓ Correct
C. $\dfrac{\mu_0I}{R}$
D. $\dfrac{\mu_0I}{2\pi R}$
Solution: Every element of the loop is at the same distance $R$ and contributes in the same direction along the axis.
Q4 — Biot-Savart Law & Field due to Current · medium · theory
A straight conductor carrying current $I$ is bent into a semicircle of radius $R$. The field at the centre of curvature is:
A. $\dfrac{\mu_0I}{2R}$
B. $\dfrac{\mu_0I}{8R}$
C. $\dfrac{\mu_0I}{2\pi R}$
D. $\dfrac{\mu_0I}{4R}$ ✓ Correct
Solution: A semicircle subtends half the full turn, so it gives half the full-circle value $\dfrac{\mu_0I}{2R}$.
Q5 — Biot-Savart Law & Field due to Current · easy · theory
The direction of the magnetic field around a straight current-carrying wire is given by:
A. Lenz's law
B. The right-hand thumb rule ✓ Correct
C. Fleming's left-hand rule
D. Coulomb's law
Solution: With the thumb along the current, the curled fingers give the sense in which the circular field lines run.
Q6 — Biot-Savart Law & Field due to Current · medium · theory
The value of the permeability of free space $\mu_0$ is:
A. $4\pi \times 10^{-7}\text{ T}\cdot\text{m}/\text{A}$ ✓ Correct
B. $9 \times 10^9\text{ T}\cdot\text{m}/\text{A}$
C. $4\pi \times 10^{-11}\text{ T}\cdot\text{m}/\text{A}$
D. $8.85 \times 10^{-12}\text{ T}\cdot\text{m}/\text{A}$
Solution: It appears in the Biot-Savart law and gives $\dfrac{\mu_0}{4\pi} = 10^{-7}$ in SI units.
Q7 — Biot-Savart Law & Field due to Current · easy · theory
The magnetic field at the centre of a circular coil of $N$ turns is:
A. $\dfrac{\mu_0NI}{2R}$ ✓ Correct
B. $\dfrac{\mu_0I}{2NR}$
C. $\dfrac{\mu_0NI}{2\pi R}$
D. $\dfrac{\mu_0NI}{4R}$
Solution: Each turn contributes equally, so the single-turn result is simply multiplied by $N$.
Q8 — Biot-Savart Law & Field due to Current · hard · theory
The magnetic field on the axis of a circular coil, far from the coil, falls off as:
A. $\dfrac{1}{x}$
B. $\dfrac{1}{x^2}$
C. $\dfrac{1}{x^4}$
D. $\dfrac{1}{x^3}$ ✓ Correct
Solution: At large distances the loop behaves as a magnetic dipole, whose axial field varies inversely as the cube of the distance.
Q9 — Biot-Savart Law & Field due to Current · easy · theory
The SI unit of magnetic field induction is the:
A. Tesla ✓ Correct
B. Gauss
C. Henry
D. Weber
Solution: One tesla is one weber per square metre; the gauss is the CGS unit, with $1\text{ T} = 10^4\text{ G}$.
Q10 — Biot-Savart Law & Field due to Current · medium · numerical
A current of $5\text{ A}$ flows through a long straight wire. The field at $10\text{ cm}$ from it is:
A. $2 \times 10^{-5}\text{ T}$
B. $10^{-6}\text{ T}$
C. $10^{-5}\text{ T}$ ✓ Correct
D. $5 \times 10^{-6}\text{ T}$
Solution: $B = \dfrac{\mu_0I}{2\pi r} = \dfrac{2 \times 10^{-7} \times 5}{0.1} = 10^{-5}\text{ T}$.
Q11 — Biot-Savart Law & Field due to Current · medium · numerical
A long straight wire carries $10\text{ A}$. The field at $20\text{ cm}$ from it is:
A. $5 \times 10^{-6}\text{ T}$
B. $4 \times 10^{-5}\text{ T}$
C. $10^{-5}\text{ T}$ ✓ Correct
D. $2 \times 10^{-5}\text{ T}$
Solution: $B = \dfrac{2 \times 10^{-7} \times 10}{0.2} = 10^{-5}\text{ T}$.
Q12 — Biot-Savart Law & Field due to Current · hard · numerical
A circular coil of radius $0.1\text{ m}$ carries a current of $2\text{ A}$. The field at its centre is approximately:
A. $2.5 \times 10^{-5}\text{ T}$
B. $1.26 \times 10^{-6}\text{ T}$
C. $4 \times 10^{-6}\text{ T}$
D. $1.26 \times 10^{-5}\text{ T}$ ✓ Correct
Solution: $B = \dfrac{\mu_0I}{2R} = \dfrac{4\pi \times 10^{-7} \times 2}{0.2} \approx 1.26 \times 10^{-5}\text{ T}$.
Q13 — Biot-Savart Law & Field due to Current · hard · numerical
A circular coil of radius $0.2\text{ m}$ carries $4\text{ A}$. The field at its centre is approximately:
A. $1.26 \times 10^{-5}\text{ T}$ ✓ Correct
B. $6.3 \times 10^{-6}\text{ T}$
C. $2.5 \times 10^{-5}\text{ T}$
D. $5 \times 10^{-5}\text{ T}$
Solution: $B = \dfrac{4\pi \times 10^{-7} \times 4}{2 \times 0.2} \approx 1.26 \times 10^{-5}\text{ T}$.
Q14 — Biot-Savart Law & Field due to Current · medium · numerical
A circular coil carries current $I$. If both the current and the radius are doubled, the field at the centre:
A. Becomes four times
B. Halves
C. Remains unchanged ✓ Correct
D. Doubles
Solution: $B = \dfrac{\mu_0I}{2R}$ depends on the ratio $\dfrac{I}{R}$, which is unaltered when both are doubled.
Q15 — Biot-Savart Law & Field due to Current · hard · numerical
A coil of $100$ turns and radius $0.05\text{ m}$ carries $1\text{ A}$. The field at its centre is approximately:
A. $2.5 \times 10^{-3}\text{ T}$
B. $6.3 \times 10^{-4}\text{ T}$
C. $1.26 \times 10^{-5}\text{ T}$
D. $1.26 \times 10^{-3}\text{ T}$ ✓ Correct
Solution: $B = \dfrac{\mu_0NI}{2R} = \dfrac{4\pi \times 10^{-7} \times 100 \times 1}{0.1} \approx 1.26 \times 10^{-3}\text{ T}$.
Q16 — Biot-Savart Law & Field due to Current · easy · numerical
If the current in a long straight wire is doubled, the field at a fixed point near it:
A. Remains unchanged
B. Doubles ✓ Correct
C. Becomes four times
D. Halves
Solution: $B = \dfrac{\mu_0I}{2\pi r} \propto I$ at a fixed distance.
Q17 — Biot-Savart Law & Field due to Current · easy · numerical
If the distance from a long straight current-carrying wire is doubled, the field there:
A. Becomes one-fourth
B. Doubles
C. Remains unchanged
D. Halves ✓ Correct
Solution: $B \propto \dfrac{1}{r}$, so doubling the distance halves the field.
Q18 — Biot-Savart Law & Field due to Current · medium · numerical
A long straight wire carries $2\text{ A}$. The field at $5\text{ cm}$ from it is:
A. $1.6 \times 10^{-5}\text{ T}$
B. $2 \times 10^{-6}\text{ T}$
C. $8 \times 10^{-6}\text{ T}$ ✓ Correct
D. $4 \times 10^{-6}\text{ T}$
Solution: $B = \dfrac{2 \times 10^{-7} \times 2}{0.05} = 8 \times 10^{-6}\text{ T}$.
Q19 — Biot-Savart Law & Field due to Current · hard · numerical
A coil of $50$ turns and radius $0.1\text{ m}$ carries $2\text{ A}$. The field at its centre is approximately:
A. $1.26 \times 10^{-4}\text{ T}$
B. $3.1 \times 10^{-4}\text{ T}$
C. $6.3 \times 10^{-4}\text{ T}$ ✓ Correct
D. $1.26 \times 10^{-3}\text{ T}$
Solution: $B = \dfrac{4\pi \times 10^{-7} \times 50 \times 2}{0.2} \approx 6.3 \times 10^{-4}\text{ T}$.
Q20 — Biot-Savart Law & Field due to Current · hard · numerical
A conductor carrying current $I$ is bent into a quarter circle of radius $R$. The field at the centre of curvature is:
A. $\dfrac{\mu_0I}{4R}$
B. $\dfrac{\mu_0I}{2R}$
C. $\dfrac{\mu_0I}{8R}$ ✓ Correct
D. $\dfrac{\mu_0I}{16R}$
Solution: A quarter circle gives one-fourth of the full-circle field $\dfrac{\mu_0I}{2R}$.
Q21 — Biot-Savart Law & Field due to Current · medium · numerical
Two long straight parallel wires carry equal currents in the same direction. Midway between them the resultant magnetic field is:
A. Zero ✓ Correct
B. Maximum
C. Equal to that of one wire
D. Twice that of one wire
Solution: At the midpoint the two fields are equal in magnitude but oppositely directed, so they cancel.