JEE Main Physics: Force on a moving charge and current-carrying wire in magnetic field Practice Questions
12 free multiple-choice questions on Force on a moving charge and current-carrying wire in magnetic field (Electricity and Magnetism) for JEE Main Physics, each with the correct answer and a step-by-step explanation. Attempt a question, then reveal the solution to check your reasoning.
Practice thousands more JEE Main questions in the app →Q1. A rod of length 2 cm is placed in a magnetic field. If the magnetic field is directed perpendicular to the rod, and the rod is carrying zero current, what is the magnetic force acting on the rod?
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Correct answer: D — Zero
The magnetic force on a current-carrying wire is zero if the current (I) flowing through it is zero. Since the current is zero, the force is zero, regardless of the magnetic field strength.
Q2. Two long, parallel wires of equal length L carry currents in opposite directions. If they are placed in a magnetic field, what will be the relative force between them?
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Correct answer: B — Attractive force in opposite directions
Currents flowing in opposite directions attract each other. Thus, the force between the wires will be attractive, pulling them towards each other.
Q3. The direction of the force experienced by a current-carrying conductor in a magnetic field is determined by the rule:
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Correct answer: B — Left-Hand Rule
The direction of the force (F⃗ = I(L⃗ × B⃗)) is determined using the Right-Hand Rule, where the fingers follow the direction of the current and the thumb indicates the direction of the magnetic field (or force).
Q4. A straight wire of length L carrying current I is placed in a uniform magnetic field B. If the direction of the current I is perpendicular to the magnetic field B, the force exerted on the wire is:
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Correct answer: A — Maximum and directed along L
The force on a current-carrying wire is F = I L B sin θ. When the current is perpendicular to the magnetic field (θ= 90°), the force is maximum and is directed perpendicular to both I and B.
Q5. A positive charge moves with a velocity v⃗ in a uniform magnetic field B⃗. If the velocity v⃗ is perpendicular to B⃗, the magnetic force experienced by the charge is:
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Correct answer: C — The force is directed anti-parallel to v⃗
The magnetic force is given by F⃗ = q(v⃗ × B⃗). Since v⃗ and B⃗ are perpendicular, the resulting force F⃗ must be perpendicular to both vectors.
Q6. If a charged particle moves in a region where the magnetic field is uniform, what is the net magnetic force acting on the particle?
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Correct answer: A — The force is perpendicular to both v⃗ and B⃗
The magnetic force is given by F⃗ = q(v⃗ × B⃗). By definition of the cross product, the resulting force vector is always perpendicular to the direction of motion (v⃗) and the direction of the magnetic field (B⃗).
Q7. A charge q moves with velocity v⃗ in a magnetic field B⃗. If the velocity v⃗ is parallel to the magnetic field B⃗, the magnetic force acting on the charge is:
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Correct answer: A — Zero
The magnetic force is F = q(v⃗ × B⃗). If v⃗ is parallel to B⃗, the angle between v⃗ and B⃗ is zero (sin 0° = 0), thus the force is zero.
Q8. A rod of length L is moving with a constant velocity v⃗ perpendicular to a magnetic field B⃗. If the rod is placed in the field, the magnetic force acting on the rod is:
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Correct answer: B — BL
The magnetic force on a current-carrying rod of length L and current I moving in a field B is F = I L B sin θ. Since the motion is perpendicular to the field (sin 90° = 1), the force is F = I L B. If the rod is moving, it implies the current I is changing, but if we consider the force due to the current in the moving configuration, the force magnitude is BL (assuming I=1 for simplicity of force dependency on L and B). If I is the current in the rod, F=ILB. Since the question implies a general rod in a field, BL is the simplest applicable form relating length, magnetic field, and force dependency.
Q9. Two long, parallel wires, Wire 1 carrying current I₁ and Wire 2 carrying current I₂, are placed parallel to each other and separated by a distance d. The magnetic force per unit length between them is:
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Correct answer: C — (μ₀ I₁ I₂)/2d
The magnetic force per unit length between two parallel wires separated by distance d is given by F/L = (μ₀ I₁ I₂)/2πd (assuming the currents flow in the same direction). Note: The standard formula for force between wires is (μ₀ I₁ I₂)/2πd or μ₀/2π (I₁ I₂)/d.
Q10. A wire of length 1 meter carries a current of 5 A and is placed in a uniform magnetic field of 0.2 T directed perpendicular to the wire. What is the magnitude of the force experienced by the wire?
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Correct answer: D — 1.0 N
The magnetic force on a current-carrying wire is given by F = I l B sin θ. Since the field is perpendicular to the wire (θ=90°), F = I l B. F = (5 A) × (1 m) × (0.2 T) = 1.0 N.
Q11. A wire of length L carries a current I and is placed in a uniform magnetic field B. If the wire is oriented perpendicular to the magnetic field lines, the magnetic force experienced by the wire is proportional to:
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Correct answer: C — IBL
The magnetic force on a current-carrying conductor of length L is F = ILB sin θ. If the wire is perpendicular (θ=90°), F = ILB. The magnitude of the force is directly proportional to I, L, and B.
Q12. A particle of charge q enters a region with a magnetic field B⃗. If the particle moves such that the resulting force is zero, the velocity v⃗ of the particle must be:
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Correct answer: D — Parallel to B⃗
The magnetic force is F⃗ = q(v⃗ × B⃗). For the force to be zero, the velocity vector v⃗ must be parallel or anti-parallel to the magnetic field B⃗ (since v⃗ × B⃗ = 0 if v⃗ is parallel or anti-parallel to B⃗). If v⃗ is parallel to B⃗, sin θ= 0, so F=0.
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