+91-8076753736 [email protected]

-->

Moving Charges And Magnetism

Question

Wired Faculty App

Three long straight and parallel wires, carrying currents, are arranged as shown in the figure below. Find the force experienced by a 25 cm length of wire C.

Solution

Length of the wire on which force is experienced, l = 25 cm
As seen in the fig. the direction of current in C and D wires are in opposite direction so, force of repulsion F will act on C towards right side. Similarly, due to G it will be in left hand side so resultant force on C due to wires D and G will be F.
Thus,

F = F_{1} - F_{2} = \frac{μ_{0} I_{1} I_{2} l}{2 {πr}_{1}} - \frac{μ_{0} I_{1} I_{3} l}{2 {πr}_{2}}

= \frac{μ_{0} I_{1} l}{2 π} [\frac{I_{2}}{r_{1}} - \frac{I_{3}}{r_{2}}]

= \frac{4 π \times 10^{- 7} \times 10 \times 25}{2 π} [\frac{30}{0.03} - \frac{20}{0.10}] = 5 \times 10^{- 7} \times 80 = 400 \times 10^{- 7}

F_c = 4 x 10^–5 N towards left side.

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation