B)-0.325 V C)-0.100 v D) 0.100 V E) 0.325 V 11. The 0.20 m-long rod in the figur

Question

B)-0.325 V C)-0.100 v D) 0.100 V E) 0.325 V 11. The 0.20 m-long rod in the figure is moving towards east at a rate of 25.0 m/s in a magnetic field of of 0.30 T (vertically downward). The electromotive-force (emf) generated by the moving rod is: A) 1.5 V, with clockwise current B) 1.5 V, with counter-clockwise current C) 15 V, with clockwise current D) 15 V, with counter-clockwise current E) zero V 12. In problem 11, the electric field in the rod: A) zero V/m Bis downward , B) 7.50 V/m, towards south C) 7.50 V/m, towards north D) 37.5 V/m, towards north E) 37.5 V/m, towards south 13. You are designing a generator to provide a rms voltage of 120.0 V. If the ge s coil has 40.00 turns cross-sectional area of 0.08500 m2, what would be the frequency of the generator in a magnetic field of 0.15 A) 60 Hz B) 50 Hz C) 40 Hz D)30? E) 20 Hz 14. The coil of a generator has 24 loops and a cross-sectional area of 0.50 m2, What is the maximum emf gener by this generator if it is spinning with an angular velocity of 7.9 rad/s in a 2.1 T magnetic field? A) 50 V B) 100 V C) 200 V D) 300 V E) 400 V

Explanation / Answer

11.

Emf induced,

e= BLv

= 0.3*0.2*25

= 1.5 V

Using Fleming's right hand rule, direction of current is anticlockwise in the loop.

-------------------------

12.

Electric field in the rod

E= V/d= 1.5/0.2

E= 7.5 V/m, to the north.

- - - - - - - - - - - - - - - -------

13

Max enf induced

Eo= NBA (2 *pi*f)

120sqrt2= 40*0.159*0.085*2*3.14*f

f= 50 Hz.

-----------------

14

Eo= NBAw

= 24*0.5*7.9*2.1

=

------------

Do comment in case any doubt.. Goodluck

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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