A current of 3 , A is flowing in a plane circ | vedclass

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(B) The magnetic dipole moment $M$ of a current-carrying coil is given by the formula:$M = N \cdot I \cdot A$Where:$N = 20$ (number of turns)$I = 3 \,

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$0.3$

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(B) The magnetic dipole moment $M$ of a current-carrying coil is given by the formula:$M = N \cdot I \cdot A$Where:$N = 20$ (number of turns)$I = 3 \, A$ (current)$A = \pi \cdot r^2$ (area of the coil)Given radius $r = 4 \, cm = 0.04 \, m$.Substituting the values:$A = \pi \cdot (0.04)^2 = \pi \cdot 0.0016 \, m^2$$M = 20 	imes 3 	imes \pi 	imes 0.0016$$M = 60 	imes 0.0050265$$M \approx 0.3016 \, A \cdot m^2$Rounding to the nearest provided option,the value is $0.3 \, A \cdot m^2$.

$A$ current of $3 \, A$ is flowing in a plane circular coil of radius $4 \, cm$ and number of turns $20$. The coil is placed in a uniform magnetic field of magnetic induction $0.5 \, T$. The magnetic dipole moment of the coil is $... \, A \cdot m^2$.

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