CBSE 2025 · Region 2 · Set 1 · Q32 · 5 marks
(i)What is the source of force acting on a current-carrying conductor placed in a magnetic field? Obtain the expression for force acting between two long straight parallel conductors carrying steady currents and hence define 'ampere'.(ii)A point charge q is moving with velocity $\displaystyle \vec{v}$ in a uniform magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$. Find the work done by the magnetic force on the charge.(iii)Explain the necessary conditions in which the trajectory of a charged particle is helical in a uniform magnetic field.(i)A current carrying loop can be considered as a magnetic dipole placed along its axis. Explain.(ii)Obtain the relation for magnetic dipole moment $\displaystyle \overrightarrow{\mathrm{M}}$ of current carrying coil. Give the direction of $\displaystyle \overrightarrow{\mathrm{M}}$.(iii)A current carrying coil is placed in an external uniform magnetic field. The coil is free to turn in the magnetic field. What is the net force acting on the coil? Obtain the orientation of the coil in stable equilibrium. Show that in this orientation the flux of the total field (field produced by the loop + external field) through the coil is maximum.
(i)
What is the source of force acting on a current-carrying conductor placed in a magnetic field? Obtain the expression for force acting between two long straight parallel conductors carrying steady currents and hence define 'ampere'.
(ii)
A point charge q is moving with velocity $\displaystyle \vec{v}$ in a uniform magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$. Find the work done by the magnetic force on the charge.
(iii)
Explain the necessary conditions in which the trajectory of a charged particle is helical in a uniform magnetic field.
(i)
A current carrying loop can be considered as a magnetic dipole placed along its axis. Explain.
(ii)
Obtain the relation for magnetic dipole moment $\displaystyle \overrightarrow{\mathrm{M}}$ of current carrying coil. Give the direction of $\displaystyle \overrightarrow{\mathrm{M}}$.
(iii)
A current carrying coil is placed in an external uniform magnetic field. The coil is free to turn in the magnetic field. What is the net force acting on the coil? Obtain the orientation of the coil in stable equilibrium. Show that in this orientation the flux of the total field (field produced by the loop + external field) through the coil is maximum.
Marking-scheme solution
(i)
The source of force is the interaction between the field produced by the current carrying conductor and the external field in which it is placed.
Two long parallel conductors a & b, separated by a distance d, carrying currents $\displaystyle I_{a}$ and $\displaystyle I_{b}$, respectively.
The magnetic field due to a:
$\displaystyle B_{a}=\frac{\mu_{0} I_{a}}{2 \pi d}$
The force $\displaystyle F_{b a}$ is the force on a segment L of 'b' due to 'a'.
$\displaystyle F_{b a}=I_{b} L B_{a}$
$\displaystyle =\frac{\mu_{0} I_{a} I_{b}}{2 \pi d} L$
Definition – The 'ampere' is that value of steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to $\displaystyle 2 \times 10^{-7}$ newton per metre of length.
(ii)
Work done by the magnetic force on the charge is zero as force is perpendicular to $\displaystyle \vec{v}$.
(iii)
The velocity $\displaystyle (\vec{v})$ is at an arbitrary angle $\displaystyle \theta$ w.r.t the magnetic field $\displaystyle (\vec{B})$.
(i)
The two faces of a current carrying loop behave like two poles of a magnet therefore can be considered as a magnetic dipole placed along its axis.
(ii)
Magnetic moment (M) $\displaystyle \propto$ Current (I)
$\displaystyle \propto$ Area (A)
$\displaystyle \therefore M=I A$
Direction is same as the area vector.
(iii)
Net force acting on the coil is zero.
The potential energy $\displaystyle \left(U_{B}\right)$ of a current carrying loop in an external magnetic field $\displaystyle =-\vec{M} \cdot \vec{B}$
For the coil to be in stable equilibrium $\displaystyle U_{B}$ should be minimum so $\displaystyle \theta=0^{\circ}$.
Therefore, magnetic flux $\displaystyle (\phi)$ due to the total field $\displaystyle =\left(B_{\text {coil }}+B_{\text {ext }}\right) A$, which is its maximum value.
Moving Charges and MagnetismForce between Two Parallel Currents, the AmpereApplylong_answerhard
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.