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Physics · 2025 · 3 marks
CBSE 2025 · Region 1 · Set 2 · Q23
(a) Define magnetic moment of a current-carrying coil. Write its SI unit. (b) A coil of 60 turns and area 1.5 × 10^-3 m^3 carrying 2A current lies in…
(a)
Define magnetic moment of a current-carrying coil. Write its SI unit.
(b)
A coil of $\displaystyle 60$ turns and area $\displaystyle 1.5 \times 10^{-3} \mathrm{~m}^{3}$ carrying 2A current lies in a vertical plane. It experiences a torque of $\displaystyle 0.12$ Nm when placed in a uniform horizontal magnetic field. The torque acting on the coil changes to $\displaystyle 0.05$ Nm after the coil is rotated about its diameter by $\displaystyle 90^{\circ}$, in the magnetic field. Find the magnitude of the magnetic field.
Marking-scheme solution
(a)
Magnetic moment of a current carrying coil is defined as the product of current flowing through the coil and area of the coil.
$\displaystyle M=I A$
S.I. unit is $\displaystyle \mathrm{Am}^{2}$
(b)
$\displaystyle \tau=N I A B \sin \theta$
$\displaystyle \tau_{1}=0.12=60 \times 2 \times 1.5 \times 10^{-3} \times B \sin \theta$
$\displaystyle B \sin \theta=\frac{2}{3}$
$\displaystyle \tau_{2}=0.05=60 \times 2 \times 1.5 \times 10^{-3} \times B \cos \theta$
$\displaystyle B \cos \theta=\frac{5}{18}$
$\displaystyle B=\sqrt{B^{2} \sin ^{2} \theta+B^{2} \cos ^{2} \theta}$
$\displaystyle =\sqrt{\left(\frac{2}{3}\right)^{2}+\left(\frac{5}{18}\right)^{2}}=\frac{13}{18} \mathrm{~T}$
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