A coil of area A- turns N and carrying current i is placed with its face parallel to the lines of magnetic induction B- The work done in rotating the coil through an angle of 180-0- is2iNABdfrac-iNAB-2-ZeroiNAB

Question

Toppr Admin · Accepted Answer

The potential energy of a system of magnetic moment -x2192-M placed in magnetic field -x2192-B is U-x2212-x2192-M-x2192-BHence the total work done in rotating the coil by 180-x2218- isU2-x2212-U1-MB-x2212-x2212-MB-2MB-2iNAB

A coil of area $A$ , turns $N$ and carrying current $i$ is placed with its face parallel to the lines of magnetic induction $B$ . The work done in rotating the coil through an angle of $180^{0}$ is
$2 i N A B$
$\frac{i N A B}{2}$
$Z e r o$
$i N A B$

The potential energy of a system of magnetic moment $\to M$ placed in magnetic field $\to B$ is $U = - \to M . \to B$
Hence the total work done in rotating the coil by $180^{\circ}$ is
$U_{2} - U_{1} = (M B) - (- M B)$
$= 2 M B = 2 i N A B$

A coil of area A, turns N and carrying current i is placed with its face parallel to the lines of magnetic induction B. The work done in rotating the coil through an angle of 1800 is2iNABiNAB2ZeroiNAB

The potential energy of a system of magnetic moment →M placed in magnetic field →B is U=−→M.→BHence the total work done in rotating the coil by 180∘ isU2−U1=(MB)−(−MB)=2MB=2iNAB

A coil of area $A$ , turns $N$ and carrying current $i$ is placed with its face parallel to the lines of magnetic induction $B$ . The work done in rotating the coil through an angle of $180^{0}$ is
$2 i N A B$
$\frac{i N A B}{2}$
$Z e r o$
$i N A B$

The potential energy of a system of magnetic moment $\to M$ placed in magnetic field $\to B$ is $U = - \to M . \to B$
Hence the total work done in rotating the coil by $180^{\circ}$ is
$U_{2} - U_{1} = (M B) - (- M B)$
$= 2 M B = 2 i N A B$