Name Anisha Sinih Exam #2 I. A current-carrying wire of length 50 cm is position

Question

Name Anisha Sinih Exam #2 I. A current-carrying wire of length 50 cm is positioned perpendicular to a uniform magnetic field. f c that there is a resultant force of 3.0 N on the wire due to the the current is 10.0 A and it is determined interaction of the current and field, what is the magnetic field strength? 2. There is a current I flowing in a clockwise direction in a square loop of wire (si plane of the paper. A uniform magnetic field B is toward the right (perpendicular to the left and right sides of the square loop). What is the net magnetic force acting on the loop? ide L) that is in the a) zero b) 2ILB c) 4ILB d) None of these. a long straight wire creates a magnetic field of 05 T at a distance r from the wire. The current is then doubled and the distance halved. The field strength at this new point is a) 0.1 T b) 0.2T c) unchanged d) None of these. 4. A 10-turn square coil of area 0.036 m2 and a 20-turn circular coil are both placed perpendicular to the same changing magnetic field. The voltage induced in each of the coils is the same. What is the area of the circular coil? 5. A vertically placed coaxial cable of outside diameter 1 cm has a center current of 5 amps direc upward and 4 amps on the outside cylinder directed downward. What is the B field 0.5 m from the of the cable.

Explanation / Answer

3) B = uoI/2pi*r

when the current is doubled and distance is halved

B1 / B2 = (I1/I2) * (R2 / R1)

0.05/B2 = (1/2) *(1/2)

B2 = 0.2 T

so option b) 0.2 T is correct

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