This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 2Ωresistor: I = 12V 4Ω = 3A . (b) No current through capacitor: I = 12V 6Ω = 2A . tsl353 – p.3/4 Intermediate Exam II: Problem #4 (Spring ’06) A current loop in the form of a right triangle is placed in a uniform magnetic field of magnitude B = 30 mT as shown. The current in the loop is I = 0 . 4 A in the direction indicated. (a) Find magnitude and direction of the force ~ F 1 on side 1 of the triangle. (b) Find magnitude and direction of the force ~ F 2 on side 2 of the triangle. 20cm 20cm 3 1 2 B Solution: (a) ~ F 1 = I ~ L × ~ B = 0 (angle between ~ L and ~ B is 180 ◦ ). (b) F 2 = ILB = (0 . 4A)(0 . 2m)(30 × 103 T) = 2 . 4 × 103 N . Direction of ~ F 2 : ⊗ (into plane). tsl354 – p.4/4...
View
Full Document
 Spring '07
 AndrevanTonder
 Charge, Magnetic Field, Electric charge, Intermediate Exam

Click to edit the document details