A long current carrying wire applies F force on current carrying finite wire as shown in figure $ℓ$ is length of finite wire
List-I
List-II
(I)
(P) 2F
(II)
(Q) $5 F /4$
(III)
(R) F
(IV)
(S) 0

(T) 3F

Question

A long current carrying wire applies F force on current carrying finite wire as shown in figure ℓ is length of finite wire List-I List-II (I) (P) 2F (II) (Q) 5 F /4 (III) (R) F (IV) (S) 0 (T) 3F

TestHub · Accepted Answer

Force on a finite wire parallel to a long wire: $F = \frac{μ _{0} I _{1} I _{2}}{2 π} ℓ n (\frac{r _{2}}{r _{1}})$ where r1, r2 are the distances of the near and far ends from the long wire.

Reference case: $F = \frac{μ _{0} I _{1} I _{2}}{2 π} ℓ n 2$ (l = a at distance a).

(III) l = a at distance a with current 2I2: force = 2F → III → P.

(IV) l = 6a at distance 2a: $F = \frac{μ _{0} I _{1} I _{2}}{2 π} ℓ n \frac{8 a}{2 a} = \frac{μ _{0} I _{1} I _{2}}{2 π} ℓ n 4 = 2 F$ → IV → P.

(II) Bent wire (45°-45°, l = 4√2 a) at distance 2a: net force = 2F → II → P.

(I) Wire l = 5a inclined at 37° at distance 4a: net force = 5F/4 → I → Q.

Hence I → Q; II → P; III → P; IV → P, option (A).

Physics - Magnetism Question with Solution | TestHub

Options:

Doubts & Discussion