Force on Current-Carrying Conductor — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Force on Current-Carrying Conductor MCQs with step-by-step solutions (21 questions). Part of Magnetic Fields due to Electric Current. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — 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.
Q2 — 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.
Q3 — 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.
Q4 — Force on Current-Carrying Conductor · medium · theory
Two long parallel conductors carrying currents in the same direction:
A. Exert a torque but no force
B. Exert no force
C. Repel each other
D. Attract each other ✓ Correct
Solution: Each wire sits in the field of the other, and the resulting forces pull them together when the currents are parallel.
Q5 — Force on Current-Carrying Conductor · medium · theory
The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by $d$ is:
A. $\dfrac{\mu_0I_1I_2}{4\pi d}$
B. $\dfrac{\mu_0I_1I_2}{2\pi d}$ ✓ Correct
C. $\dfrac{\mu_0I_1I_2}{2\pi d^2}$
D. $\dfrac{\mu_0I_1I_2 d}{2\pi}$
Solution: This expression is the basis of the SI definition of the ampere.
Q6 — Force on Current-Carrying Conductor · hard · theory
One ampere is defined as the current which, flowing in two infinitely long parallel wires one metre apart in vacuum, produces a force per unit length of:
A. $1\text{ N/m}$
B. $10^{-7}\text{ N/m}$
C. $4\pi \times 10^{-7}\text{ N/m}$
D. $2 \times 10^{-7}\text{ N/m}$ ✓ Correct
Solution: Substituting $I_1 = I_2 = 1\text{ A}$ and $d = 1\text{ m}$ into the force formula gives exactly this value.
Q7 — Force on Current-Carrying Conductor · medium · theory
Two long parallel conductors carrying currents in opposite directions exert on each other a force that is:
A. Attractive and proportional to $\dfrac{1}{d}$
B. Attractive and proportional to $\dfrac{1}{d^2}$
C. Repulsive and proportional to $\dfrac{1}{d}$ ✓ Correct
D. Repulsive and proportional to $\dfrac{1}{d^2}$
Solution: Antiparallel currents repel, and the force per unit length falls off inversely with the separation.
Q8 — 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.
Q9 — 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: $F = BIL = 0.5 \times 2 \times 0.5 = 0.5\text{ N}$.
Q10 — Force on Current-Carrying Conductor · medium · numerical
A wire of length $1\text{ m}$ carrying $5\text{ A}$ makes $30^\circ$ with a field of $0.2\text{ T}$. The force on it is:
A. $0.87\text{ N}$
B. $0.25\text{ N}$
C. $1\text{ N}$
D. $0.5\text{ N}$ ✓ Correct
Solution: $F = BIL\sin 30^\circ = 0.2 \times 5 \times 1 \times 0.5 = 0.5\text{ N}$.
Q11 — Force on Current-Carrying Conductor · medium · numerical
Two long parallel wires each carrying $1\text{ A}$ are $1\text{ m}$ apart. The force per unit length between them is:
A. $10^{-7}\text{ N/m}$
B. $4\pi \times 10^{-7}\text{ N/m}$
C. $2 \times 10^{-5}\text{ N/m}$
D. $2 \times 10^{-7}\text{ N/m}$ ✓ Correct
Solution: $f = \dfrac{\mu_0I_1I_2}{2\pi d} = \dfrac{2 \times 10^{-7} \times 1 \times 1}{1} = 2 \times 10^{-7}\text{ N/m}$.
Q12 — Force on Current-Carrying Conductor · hard · numerical
Two parallel wires carrying $2\text{ A}$ and $3\text{ A}$ are $10\text{ cm}$ apart. The force per unit length between them is:
A. $1.2 \times 10^{-5}\text{ N/m}$ ✓ Correct
B. $6 \times 10^{-6}\text{ N/m}$
C. $1.2 \times 10^{-6}\text{ N/m}$
D. $2.4 \times 10^{-5}\text{ N/m}$
Solution: $f = \dfrac{2 \times 10^{-7} \times 2 \times 3}{0.1} = \dfrac{1.2 \times 10^{-6}}{0.1} = 1.2 \times 10^{-5}\text{ N/m}$.
Q13 — 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.
Q14 — 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: $F = BIL = 0.1 \times 10 \times 2 = 2\text{ N}$.
Q15 — Force on Current-Carrying Conductor · medium · numerical
If both currents in a pair of parallel wires are doubled at constant separation, the force per unit length becomes:
A. Twice
B. Half
C. Four times ✓ Correct
D. Unchanged
Solution: $f \propto I_1I_2$, so doubling both multiplies the force by $4$.
Q16 — Force on Current-Carrying Conductor · medium · numerical
A conductor of length $0.5\text{ m}$ in a field of $0.2\text{ T}$ experiences a force of $1\text{ N}$ when placed perpendicular to the field. The current is:
A. $10\text{ A}$ ✓ Correct
B. $1\text{ A}$
C. $5\text{ A}$
D. $0.1\text{ A}$
Solution: $I = \dfrac{F}{BL} = \dfrac{1}{0.2 \times 0.5} = 10\text{ A}$.
Q17 — 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: $F = BIL = 0.5 \times 4 \times 1 = 2\text{ N}$.
Q18 — Force on Current-Carrying Conductor · hard · numerical
Two long parallel wires each carrying $5\text{ A}$ are $0.2\text{ m}$ apart. The force per unit length is:
A. $2.5 \times 10^{-6}\text{ N/m}$
B. $5 \times 10^{-5}\text{ N/m}$
C. $1.25 \times 10^{-5}\text{ N/m}$
D. $2.5 \times 10^{-5}\text{ N/m}$ ✓ Correct
Solution: $f = \dfrac{2 \times 10^{-7} \times 25}{0.2} = \dfrac{5 \times 10^{-6}}{0.2} = 2.5 \times 10^{-5}\text{ N/m}$.
Q19 — Force on Current-Carrying Conductor · medium · numerical
A conductor of length $0.25\text{ m}$ carrying $3\text{ A}$ lies perpendicular to a field of $0.4\text{ T}$. The force on it is:
A. $3\text{ N}$
B. $0.3\text{ N}$ ✓ Correct
C. $0.03\text{ N}$
D. $1.2\text{ N}$
Solution: $F = BIL = 0.4 \times 3 \times 0.25 = 0.3\text{ N}$.
Q20 — Force on Current-Carrying Conductor · medium · numerical
A current-carrying wire is placed at $90^\circ$ to a magnetic field and then rotated until it is parallel to the field. The force on it:
A. Reverses direction
B. Rises from zero to a maximum
C. Stays constant
D. Falls from its maximum value to zero ✓ Correct
Solution: $F = BIL\sin\theta$ decreases from $BIL$ at $90^\circ$ to zero at $0^\circ$.
Q21 — Force on Current-Carrying Conductor · hard · numerical
Two long parallel wires carry $4\text{ A}$ and $6\text{ A}$ and are $0.4\text{ m}$ apart. The force per unit length between them is:
A. $2.4 \times 10^{-5}\text{ N/m}$
B. $1.2 \times 10^{-5}\text{ N/m}$ ✓ Correct
C. $1.2 \times 10^{-6}\text{ N/m}$
D. $6 \times 10^{-6}\text{ N/m}$
Solution: $f = \dfrac{2 \times 10^{-7} \times 24}{0.4} = \dfrac{4.8 \times 10^{-6}}{0.4} = 1.2 \times 10^{-5}\text{ N/m}$.