Question

construct a quadrilateral f u s e in which a few equals to 5.2 CM, us equals to 4.6 CM as equals to 5.8 CM, angle view equals to 30 degree and Angle s equals to 150 degree​

Answer 1

Answer:

Given : for coil A:

VA=25cm=0.25m

nA=24nA=24

IA=10AI=10A

For coil B:

rb=15cm=0.15m

nb=18nb=18

Ib=15AI=15A

Magnetic field at the center of coil x

BA=µ0 /4π*2Ia πna/ra

=10−7×2×10×3.14×24/0.25

=6.028×10^−4

=6.028×10^−4T

Magnetic field at the center of coil B

B= µ0 /4π*2Ib πnb/rb

=10−7×2×3.14×15×18

=11.30×10−4T

Hence the net magnetic field at the center

Bb−Ba=(11.30×10−4−6.028×10−4)T

=5.27×10−4T