CBSE 2026 · Region 3 · Set 1 · Q23 · 3 marks
A circular coil of $\displaystyle 30$ turns and radius $\displaystyle 8.0$ cm carrying a current of $\displaystyle 6$ A is suspended vertically in a uniform horizontal magnetic field of $\displaystyle 1.0$ T . The field lines make an angle of $\displaystyle 30^{\circ}$ with the plane of the coil. Calculate the magnitude of the external torque that must be applied to prevent the coil from turning. What would happen if the circular coil is replaced by a planar coil of irregular shape that encloses the same area, keeping other parameters unchanged ?An alpha particle (mass $\displaystyle 6.4 \times 10^{-27} \mathrm{~kg}$ and charge $\displaystyle 3.2 \times 10^{-19} \mathrm{C}$ ) having $\displaystyle 8$⋅$\displaystyle 0$ MeV energy, enters a region of a uniform magnetic field of $\displaystyle 0 \cdot 5 \mathrm{~T}$. If the field is directed perpendicular to the velocity of the particle, find the radius of the circular path described by the particle. Mention the condition under which the particle in this region (i) describes a helical path, and (ii) goes straight undeviated.
A circular coil of $\displaystyle 30$ turns and radius $\displaystyle 8.0$ cm carrying a current of $\displaystyle 6$ A is suspended vertically in a uniform horizontal magnetic field of $\displaystyle 1.0$ T . The field lines make an angle of $\displaystyle 30^{\circ}$ with the plane of the coil. Calculate the magnitude of the external torque that must be applied to prevent the coil from turning. What would happen if the circular coil is replaced by a planar coil of irregular shape that encloses the same area, keeping other parameters unchanged ?
An alpha particle (mass $\displaystyle 6.4 \times 10^{-27} \mathrm{~kg}$ and charge $\displaystyle 3.2 \times 10^{-19} \mathrm{C}$ ) having $\displaystyle 8$⋅$\displaystyle 0$ MeV energy, enters a region of a uniform magnetic field of $\displaystyle 0 \cdot 5 \mathrm{~T}$. If the field is directed perpendicular to the velocity of the particle, find the radius of the circular path described by the particle. Mention the condition under which the particle in this region (i) describes a helical path, and (ii) goes straight undeviated.
Marking-scheme solution
$\displaystyle \tau=\mathrm{NIBA} \sin \theta$
$\displaystyle =30 \times 6 \times 1 \times 3.14 \times\left(8 \times 10^{-2}\right)^{2} \times \sin 60^{\circ}$
$\displaystyle =90 \times 64 \times 3.14 \times 1.73 \times 10^{-4}$
$\displaystyle =3.13 \mathrm{~N} \mathrm{~m}$
Torque will remain the same.
$\displaystyle K.E=\dfrac{q^{2} B^{2} r^{2}}{2 m}$
$\displaystyle r=\dfrac{\sqrt{2 \times K.E. \times m}}{q B}$
$\displaystyle =\dfrac{\sqrt{2 \times 8 \times 1.6 \times 10^{-19} \times 10^{6} \times 6.4 \times 10^{-27}}}{3.2 \times 10^{-19} \times 0.5}$
$\displaystyle =\dfrac{12.8 \times 10^{-20}}{1.6 \times 10^{-19}}$
$\displaystyle =8 \times 10^{-1} \mathrm{~m}=80 \mathrm{~cm}$
(i)
Particle enters the magnetic field at an angle $\displaystyle 0^{\circ}<\theta<90^{\circ}$ (Alternatively: if a component of velocity is parallel or antiparallel to the magnetic field and another component is perpendicular to it)
(ii)
Particle moves parallel or antiparallel to the magnetic field.
Moving Charges and MagnetismTorque on Current Loop, Magnetic DipoleApplyshort_answerhard
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.