Q.A straight conductor of length 0.5 m carries a current of 4 A and is placed perpendicular to a uniform magnetic field of 0.25 T. Find the force on the conductor.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Magnetic Force on a Current-Carrying Conductor
Since an electric current is physically many moving charges (conduction electrons) travelling together, a current-carrying wire in a magnetic field experiences a force that is the sum of the tiny Lorentz forces on every individual moving charge. For a straight wire of length L carrying current I in a field B applied perpendicular to it, this sum works out (after the drift speed conveniently cancels out of the calculation) to F=BIL; more generally, with a length vector L pointing along the current, F=IL×B, valid at any angle between the wire and the field. …
Use F=BIL since the conductor is perpendicular to the field. …
Given L=0.5 m, I=4 A, B=0.25 T, θ=90∘: …
Substitute directly into F=BILsinθ (Section 4.10), using sin90∘=1 since the conductor …
- Confusing B, I, L ordering -- all three multiply directly, order does not matter, but each must be substituted in consistent SI units. …
- CBSE 2026Set ANNUAL1 markMCQQ.In a uniform magnetic field B, a conductor of length l is placed parallel to the magnetic field. When a current I is passed through the conductor, the force on the conductor will be(a) IlB(b) IB/l(c) Il/B(d) zero
›Reveal solutionSolution
The magnetic force on a current-carrying wire depends on sin(theta) between the current direction and B; when the wire is parallel to B, that force is zero.
The force on a straight current-carrying conductor of length l in a uniform magnetic field B is given by
F = BIl*sin(theta)
…
- CBSE 2025Set ANNUAL1 markMCQQ.A straight current carrying wire kept in a uniform magnetic field will experience a maximum force when it is :(a) perpendicular to the magnetic field(b) parallel to the magnetic field(c) at an angle of 45° to the magnetic field(d) at an angle of 60° to the magnetic field
›Reveal solutionSolution
The magnetic force on a current-carrying wire is F=BILsinθ, which is maximum when sinθ=1, i.e. when the wire is perpendicular to B.
The force on a straight wire of length L carrying current I in a uniform field B is
F=BILsinθ
where θ is the angle between the current direction and B.
- If the wire is parallel to B (θ=0∘), sinθ=0, so F=0. …
- CBSE 2024Set A1 markMCQQ.Which one of the following is not a unit of magnetic field? (A) tesla (B) weber/metre^2 (C) newton/ampere-metre (D) newton/ampere^2
›Reveal solutionSolution
B has units tesla = Wb/m² = N/(A·m); newton/ampere² is NOT a unit of B.
Magnetic field B can be expressed as:
- tesla (T),
- weber/metre² (Wb/m²), since 1 T = 1 Wb/m²,
- newton/(ampere·metre), from F = BIL ⇒ B = F/(IL) = N/(A·m). …
- CBSE 2024Set ANNUAL1 markMCQQ.A current carrying long erect wire is kept at an angle θ with an external uniform magnetic field. The wire experiences highest force if(a) θ = 0°(b) θ = 30°(c) θ = 60°(d) θ = 90°.
›Reveal solutionSolution
The force on a current-carrying wire in a magnetic field depends on sinθ, which is maximum (=1) at θ = 90°.
A straight current-carrying conductor of length L carrying current I, placed at angle θ to a uniform magnetic field B, experiences a force
F=BILsinθ
…
- CBSE 2023Set ANNUAL1 markMCQQ.The magnetic force F (vector) on a current carrying conductor of length l (vector) in an external magnetic field B (vector) is given by(1) (I x B) / l [I=current scalar; l and B vectors](2) (l x B) / I(3) I(l x B)(4) I^2 (l x B)
›Reveal solutionSolution
Summing the Lorentz force qv x B over all the moving charges in a straight conductor of length l carrying current I gives F = I l x B.
…
- CBSE 2023Set ANNUAL1 markQ.What is the value of force on a closed circuit in a magnetic field?
›Reveal solutionSolution
The net force on any closed current loop in a uniform field is always zero.
For a closed circuit of current I in a uniform magnetic field B, the total force is
F=I∮dl×B=I(∮dl)×B …
- CBSE 2022Set GC1 markMCQQ.Current i is flowing in a wire of length l. Wire is inclined at an angle of 30∘ with the magnetic field B W-m−2. The force on the wire due to magnetic field will be:i) iBlii) iBl/2iii) 2iBliv) 23iBl
›Reveal solutionSolution
Force on a current-carrying wire in a field is F=Bilsinθ; at θ=30∘, F=iBl/2.
The magnetic force on a straight wire of length l carrying current i at angle θ to field B is
F=Bilsinθ. …
- CBSE 2021Set A1 markMCQQ.Magnetic field of 5 tesla is equal to (A) 5 × weber/(metre)² (B) 5 × 10⁵ weber/(metre)² (C) 5 × 10² weber/(metre)² (D) 5 × 10² weber × (metre)²
›Reveal solutionSolution
1 tesla = 1 weber per square metre, so 5 T = 5 Wb/m².
Magnetic flux density (magnetic field) B has the SI unit tesla. The tesla is defined from magnetic flux Φ = B·A, so B = Φ/A, giving:
1 T=1 m2Wb
…
- CBSE 2016Set ANNUAL1 markQ.What do you mean by magnetic flux density? OR What is Q-factor?
›Reveal solutionSolution
Magnetic flux density B measures the strength of a magnetic field, defined either as flux per unit area or as force per unit (current × length).
Magnetic flux density (also simply called the magnetic field, B) at a point can be defined in two equivalent ways:
(1) As flux per unit area: B = dΦ/dA, the magnetic flux passing normally through a unit area held perpendicular to the field at that point.
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