Currents of equal magnitude flow through two long parallel wires having a separation of 1.35 cm. If the force per unit length on each of the wires is $4.76\times10^{-2}$ N, what must be I?

Currents of equal magnitude flow through two long parallel wires having a separation of 1.35 cm. If the force

Numericals · Q12

4.76\times10^{-2}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

26% · 12/46 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For two long parallel wires carrying equal current $I$ , separated by distance $d$ , the force per unit length is $\dfrac{F}{L}=\dfrac{\mu_0 I^2}{2\pi d}$ (Section 10.11, with $I_1=I_2=I$ ). Solving for $I$ : $I=\sqrt{\dfrac{F}{L}\cdot\dfrac{2\pi d}{\mu_0}}$ . Substituting $F/L=4.76\times10^{-2}$ N/m, $d=1.35\ \text{cm}=0.0135$ m, $\mu_0=4\pi\times10^{-7}$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ11 — A very long straight wire carries a current of 5.2 A. What is the magn…→Next questionQ13 — The magnetic field at a distance of 2.4 cm from a long straight wire i…

Q.Currents of equal magnitude flow through two long parallel wires having a separation of 1.35 cm. If the force per unit length on each of the wires is $4.76\times10^{-2}$ N, what must be I?

Concept understanding — Force Between Two Parallel Current-Carrying Wires

Q.Currents of equal magnitude flow through two long parallel wires having a separation of 1.35 cm. If the force per unit length on each of the wires is 4.76×10−24.76\times10^{-2}4.76×10−2 N, what must be I?

🎓Unlock everything free for 14 days

Q.Currents of equal magnitude flow through two long parallel wires having a separation of 1.35 cm. If the force per unit length on each of the wires is $4.76\times10^{-2}$ N, what must be I?

Unlock everything free for 14 days