A conducting rod of length l = 0.35 m is placed on a loop of metal wire with a r

Question

A conducting rod of length l = 0.35 m is placed on a loop of metal wire with a resistance ofi? = 5 ohm as shown. A constant magnetic field of B = 0. 45 T is applied into the page. An applied force moves the rod to the right w ith a constant speed of 6.3 m/s. What is the magnitude and direction of the induced current through the resistor? A step down transformer produces a voltage of 6.0 V across the secondary coil when the voltage across the primary coil is 120 V. What voltage appears across the primary coil of this transformer if 120 V is applied to the secondary coil?

Explanation / Answer

   a.   InducedEMF   E   =   B * v *l * sin    =   0.45 * 0.35 *6.3 * sin900   =   0.992   V       Inducedcurrent   I   =   V /R   =   0.992 /5   =   0.198   A   =   0.198   mA          Applyingright hand rule, upper end or rod will be positive, i.e. thecurrent will flow form upper end to lower or anticlockwise.    b.   VS /VP   =   NS /NP       6.0 /120   =   NS /NP   =>   NS/ NP   =   1/20    When the transformer is connected inreverse,    NS' /NP'   =   NP/NS   =   20/ 1    =>   VS'/VP'   =   N'S/ NP'   =   20/ 1       VS'   /120   =   20       Voltage acrossprimary   VS'   =   120*20   =   2400   V       6.0 /120   =   NS /NP   =>   NS/ NP   =   1/20    When the transformer is connected inreverse,    NS' /NP'   =   NP/NS   =   20/ 1    =>   VS'/VP'   =   N'S/ NP'   =   20/ 1       VS'   /120   =   20       Voltage acrossprimary   VS'   =   120*20   =   2400   V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.