A metallic wire is folded to form a square loop of side a . It carries a current i and is kept perpendicular to a uniform magnetic field. If the shape of the loop is changed from square to a circle without changing the length of the wire and current, the amount of work done in doing so is

Question

A metallic wire is folded to form a square loop of side a . It carries a current i and is kept perpendicular to a uniform magnetic field. If the shape of the loop is changed from square to a circle without changing the length of the wire and current, the amount of work done in doing so is (A) ( textitiBa)2(π + 2) (B) ( textitiBa)2(π - 2) (C) ( textitiBa)2(4 / π + 1) (D) ( textitiBa)2(1 - (4/π ))

Tardigrade Admin · Accepted Answer

W = Uf - Ui =(- M textf B ( text cos 0) texto)-(- M texti B text cos 0 texto) =iB(A texti - A textf) where A texti=a2 text,A textf=π R2=(4 a2/π ) as 2π R=4a text or textitR=(2 a/π ) â´ W=iBa2(1 - (4/π ))

Q. A metallic wire is folded to form a square loop of side $a$ . It carries a current $i$ and is kept perpendicular to a uniform magnetic field. If the shape of the loop is changed from square to a circle without changing the length of the wire and current, the amount of work done in doing so is

$(iBa)^{2} (π + 2)$

$(iBa)^{2} (π - 2)$

$(iBa)^{2} (4/ π + 1)$

$(iBa)^{2} (1 - \frac{4}{π})$

$W = U_{f} - U_{i}$
$= (- M_{f} B (cos 0)^{o}) - (- M_{i} B cos 0^{o})$
$= i B (A_{i} - A_{f})$
where $A_{i} = a^{2},A_{f} = π R^{2} = \frac{4 a ^{2}}{π}$
as $2 π R = 4 a or R = \frac{2 a}{π}$
$∴ W = i B a^{2} (1 - \frac{4}{π})$

Q. A metallic wire is folded to form a square loop of side a . It carries a current i and is kept perpendicular to a uniform magnetic field. If the shape of the loop is changed from square to a circle without changing the length of the wire and current, the amount of work done in doing so is

(iBa)2(π+2)

(iBa)2(π−2)

(iBa)2(4/π+1)

(iBa)2(1−π4​)