A wire of 6.28 m length is converted into two semicircular loops of same radius as shown in the figure. Then the magnitude of magnetic force (in newton) on the wire will be

Question

A wire of 6.28 m length is converted into two semicircular loops of same radius as shown in the figure. Then the magnitude of magnetic force (in newton) on the wire will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0710b9d4261f77d0fc93327c698690c9-.png" /> | vecFm| =B I| vecL| =B I(4 R) (∵ 2 π R=6.28 ∴ R = 1 m) =5 × 1 × 4 × 1 =20 N

Q. A wire of $6.28 m$ length is converted into two semicircular loops of same radius as shown in the figure. Then the magnitude of magnetic force (in newton) on the wire will be

$∣ ∣ F_{m} ∣ ∣ = B I ∣ L ∣$
$= B I (4 R) (∵ 2 π R = 6.28 ∴ R = 1 m)$
$= 5 \times 1 \times 4 \times 1$
$= 20 N$

Q. A wire of 6.28m length is converted into two semicircular loops of same radius as shown in the figure. Then the magnitude of magnetic force (in newton) on the wire will be

∣∣​Fm​∣∣​=BI∣L∣ =BI(4R)(∵2πR=6.28∴R=1m) =5×1×4×1 =20N

Q. A wire of $6.28 m$ length is converted into two semicircular loops of same radius as shown in the figure. Then the magnitude of magnetic force (in newton) on the wire will be

$∣ ∣ F_{m} ∣ ∣ = B I ∣ L ∣$
$= B I (4 R) (∵ 2 π R = 6.28 ∴ R = 1 m)$
$= 5 \times 1 \times 4 \times 1$
$= 20 N$