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Magnetic Fields due to Electric Current — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Magnetic Fields due to Electric Current MCQs with step-by-step solutions covering Magnetic Force & Motion of Charges, Biot-Savart Law & Field due to Current, Ampere's Circuital Law & Solenoid, Force on Current-Carrying Conductor, Torque on Current Loop & Magnetic Moment, Moving Coil Galvanometer & Conversions. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Magnetic Force & Motion of Charges · easy · theory
The magnetic force on a charge $q$ moving with velocity $v$ at an angle $\theta$ to a field $B$ is:
A. $qvB$
B. $\dfrac{qv}{B}\sin\theta$
C. $qvB\cos\theta$
D. $qvB\sin\theta$ ✓ Correct
Solution: The force is greatest when the velocity is perpendicular to the field and vanishes when the two are parallel.
Q2 — Magnetic Force & Motion of Charges · easy · theory
A stationary charge placed in a uniform magnetic field experiences:
A. A force along the field
B. No force ✓ Correct
C. A force perpendicular to the field
D. A force that increases with time
Solution: The magnetic force is proportional to velocity, so a charge at rest feels nothing from a magnetic field.
Q3 — Magnetic Force & Motion of Charges · easy · theory
A charged particle entering a uniform magnetic field perpendicular to its velocity follows:
A. A parabolic path
B. A helical path
C. A circular path ✓ Correct
D. A straight line
Solution: The constant-magnitude force always perpendicular to the velocity supplies exactly the centripetal force for uniform circular motion.
Q4 — Magnetic Force & Motion of Charges · easy · numerical
A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is:
A. Zero ✓ Correct
B. Perpendicular to the field
C. Maximum
D. Equal to $qvB$
Solution: $F = qvB\sin 0^\circ = 0$, so the particle travels undeflected in a straight line.
Q5 — Magnetic Force & Motion of Charges · easy · numerical
If the speed of a charged particle in a magnetic field is doubled, the radius of its circular path:
A. Remains unchanged
B. Halves
C. Doubles ✓ Correct
D. Becomes four times
Solution: $r = \dfrac{mv}{qB} \propto v$, so doubling the speed doubles the radius.
Q6 — Magnetic Force & Motion of Charges · easy · numerical
If the magnetic field is doubled, the radius of the circular path of a charged particle:
A. Halves ✓ Correct
B. Becomes four times
C. Doubles
D. Remains unchanged
Solution: $r \propto \dfrac{1}{B}$, so a stronger field bends the particle into a tighter circle.
Q7 — 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.
Q8 — 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.
Q9 — 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.
Q10 — 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 $N$.
Q11 — 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 $1\text{ T} = 10^4\text{ G}$.
Q12 — 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.
Q13 — 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.
Q14 — Ampere's Circuital Law & Solenoid · easy · theory
Ampere's circuital law states that:
A. $\oint \vec{B}\cdot d\vec{l} = \mu_0I_{enc}$ ✓ Correct
B. $\oint \vec{B}\cdot d\vec{A} = \mu_0I_{enc}$
C. $\oint \vec{B}\times d\vec{l} = \mu_0I_{enc}$
D. $\oint \vec{B}\cdot d\vec{l} = 0$ always
Solution: The line integral of $\vec{B}$ round any closed path equals $\mu_0$ times the net current threading that path.
Q15 — Ampere's Circuital Law & Solenoid · easy · theory
The magnetic field inside a long solenoid with $n$ turns per unit length carrying current $I$ is:
A. $\dfrac{\mu_0nI}{2}$
B. $\dfrac{\mu_0I}{2n}$
C. $\mu_0n^2I$
D. $\mu_0nI$ ✓ Correct
Solution: Applying Ampere's law to a rectangular loop partly inside the solenoid gives this uniform interior field.
Q16 — Ampere's Circuital Law & Solenoid · easy · theory
For a solenoid of $N$ turns wound over a length $L$, the number of turns per unit length is:
A. $\dfrac{N}{2L}$
B. $NL$
C. $\dfrac{N}{L}$ ✓ Correct
D. $\dfrac{L}{N}$
Solution: This quantity $n$ is what enters the expression $B = \mu_0nI$.
Q17 — Ampere's Circuital Law & Solenoid · easy · numerical
If the number of turns per unit length of a solenoid is doubled at constant current, the interior field:
A. Halves
B. Becomes four times
C. Remains unchanged
D. Doubles ✓ Correct
Solution: $B = \mu_0nI \propto n$.
Q18 — Ampere's Circuital Law & Solenoid · easy · numerical
If the current through a solenoid is doubled, the field inside it:
A. Becomes four times
B. Remains unchanged
C. Halves
D. Doubles ✓ Correct
Solution: $B = \mu_0nI \propto I$.
Q19 — Force on Current-Carrying Conductor · easy · theory
The force on a straight conductor of length $L$ carrying current $I$ in a field $B$ at angle $\theta$ is:
A. $BIL\tan\theta$
B. $\dfrac{BI}{L}\sin\theta$
C. $BIL\sin\theta$ ✓ Correct
D. $BIL\cos\theta$
Solution: It is greatest when the conductor is perpendicular to the field and vanishes when it lies along the field.
Q20 — Force on Current-Carrying Conductor · easy · theory
A straight wire carrying current is placed parallel to a uniform magnetic field. The force on it is:
A. $BIL$
B. Zero ✓ Correct
C. $\dfrac{BIL}{2}$
D. $2BIL$
Solution: $F = BIL\sin 0^\circ = 0$, since the current and field are collinear.
Q21 — Force on Current-Carrying Conductor · easy · theory
The direction of the force on a current-carrying conductor in a magnetic field is given by:
A. Fleming's left-hand rule ✓ Correct
B. Lenz's law
C. The right-hand thumb rule
D. Fleming's right-hand rule
Solution: With the forefinger along the field and the middle finger along the current, the thumb gives the force — Fleming's left-hand rule.
Q22 — Force on Current-Carrying Conductor · easy · theory
The force on a current-carrying conductor in a magnetic field is directed:
A. Along the field
B. Perpendicular to both the current and the field ✓ Correct
C. Opposite to the current
D. Along the current
Solution: $\vec{F} = I\vec{L} \times \vec{B}$ is a cross product, so it is normal to the plane containing both.
Q23 — Force on Current-Carrying Conductor · easy · numerical
A conductor of length $0.5\text{ m}$ carrying $2\text{ A}$ lies perpendicular to a field of $0.5\text{ T}$. The force on it is:
A. $0.5\text{ N}$ ✓ Correct
B. $2\text{ N}$
C. $0.25\text{ N}$
D. $1\text{ N}$
Solution: $F = BIL = 0.5 \times 2 \times 0.5 = 0.5\text{ N}$.
Q24 — Force on Current-Carrying Conductor · easy · numerical
If the separation between two current-carrying parallel wires is doubled, the force per unit length becomes:
A. Unchanged
B. One-fourth as large
C. Twice as large
D. Half as large ✓ Correct
Solution: $f \propto \dfrac{1}{d}$, so doubling the separation halves the force.
Q25 — Force on Current-Carrying Conductor · easy · numerical
A conductor of length $2\text{ m}$ carrying $10\text{ A}$ lies perpendicular to a field of $0.1\text{ T}$. The force on it is:
A. $1\text{ N}$
B. $0.5\text{ N}$
C. $20\text{ N}$
D. $2\text{ N}$ ✓ Correct
Solution: $F = BIL = 0.1 \times 10 \times 2 = 2\text{ N}$.
Q26 — Force on Current-Carrying Conductor · easy · numerical
A wire of length $1\text{ m}$ carrying $4\text{ A}$ lies perpendicular to a field of $0.5\text{ T}$. The force on it is:
A. $2\text{ N}$ ✓ Correct
B. $4\text{ N}$
C. $0.5\text{ N}$
D. $8\text{ N}$
Solution: $F = BIL = 0.5 \times 4 \times 1 = 2\text{ N}$.
Q27 — Torque on Current Loop & Magnetic Moment · easy · theory
The magnetic dipole moment of a coil of $N$ turns, area $A$ carrying current $I$ is:
A. $NIA$ ✓ Correct
B. $\dfrac{NI}{A}$
C. $N^2IA$
D. $\dfrac{NA}{I}$
Solution: Its direction is normal to the plane of the coil, given by the right-hand rule applied to the current.
Q28 — Torque on Current Loop & Magnetic Moment · easy · theory
The torque on a current loop of magnetic moment $m$ in a uniform field $B$ at angle $\theta$ is:
A. $mB\tan\theta$
B. $mB\sin\theta$ ✓ Correct
C. $mB\cos\theta$
D. $\dfrac{mB}{\sin\theta}$
Solution: Equivalently $\tau = NIAB\sin\theta$, where $\theta$ is measured between $\vec{m}$ and $\vec{B}$.
Q29 — Torque on Current Loop & Magnetic Moment · easy · theory
The SI unit of magnetic dipole moment is:
A. $\text{J}\cdot\text{T}$
B. $\text{A}\cdot\text{m}$
C. $\text{A}\cdot\text{m}^2$ ✓ Correct
D. $\text{A}/\text{m}$
Solution: From $m = IA$, the unit is ampere times square metre, equivalently joule per tesla.
Q30 — Torque on Current Loop & Magnetic Moment · easy · numerical
A dipole of magnetic moment $2\text{ A}\cdot\text{m}^2$ is placed in a field of $0.3\text{ T}$. The maximum torque on it is:
A. $0.15\text{ N}\cdot\text{m}$
B. $0.3\text{ N}\cdot\text{m}$
C. $0.6\text{ N}\cdot\text{m}$ ✓ Correct
D. $6\text{ N}\cdot\text{m}$
Solution: $\tau_{max} = mB = 2 \times 0.3 = 0.6\text{ N}\cdot\text{m}$.