Q.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
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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. …
Magnetic field strength can be expressed in several equivalent SI units derived from its defining relations, so identifying the option that does NOT match any of those equivalent forms answers the question. …
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 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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