6.67 points Save Answer Question 1 A particle (g4.0 C, 5.0 mg) moves in a unifor
ID: 1792577 • Letter: 6
Question
6.67 points Save Answer Question 1 A particle (g4.0 C, 5.0 mg) moves in a uniform magnetie field with a velocity having a magnirude of 2.0 km's ad a direction that is s0° away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s2. What is the magnitude of the magnetic field? a. 5.3 mT (i) b. 5.1 mT C. 4.7 m1T d. 4.9 mT e. 3.6 mT 6.67 points Save Answer Question 2 A straight wire of length L carries a current I in the positive - direction in a region where the magnetic field is uniform and specified by B magnetic force on the wire? 3B, B2, and B. -B, where B is a constant. What is the magnitude of the a. 4.2 ILB b. 3.2 ILB C.1.0 ILB d. 3.6 ILB e, 5.0 11.B 6.67 points Save Answer Question 3 The figure shows the orientation of a rectangular loop consisting of 80 closely wrapped turns each carrying a current I. The magnetic field in the region is (40i) mT. The loop can turn about the y axis. If 0-30. a-0.40 m. 0.30 m, and I-8.0 A, what is the magnitude of the torque exerted on the loop? a. 0.34 N·m b.3.1N C. 2.5 N-m d. 2.7N-mExplanation / Answer
The magnetic force acting on a moving charge placed in magnetic field is
F= Bqv sin theta
ma = Bqv sin theta
B = ma/ qv sin theta
= 5 * 10^-6 ( 5.8)/4 * 10^-6 * 2 * 10^3 * sin 50
=4.7 mT
(b)
the net magnetic field is
B = sqrt Bx^2 + By ^2
= sqrt ( 3B)^2 + ( 2B)^2
=3.6 B
F = BIL sin theta
= 3.6 B IL sin 90
= 3.6ILB
(3)
T= NIAB sin theta
= 80 ( 0.4) (0.3) 8 ( 40* 10^-3 ) sin30
=1.5 Nm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.