Magnetic Effects Of Current Question 427
Question: Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: $ I_{1}=20,A $ , $ I_{2}=-6A $ , $ I_{3}=12A $ , $ I_{4}=-7A $ , $ I_{5}=18A $ . (Negative currents are opposite in direction to the positive). The magnetic field induction at a distance of 10 cm from the cable is
Options:
A) $ 5\mu T $
B) $ 15\mu T $
C) $ 74\mu T $
D) $ 128\mu T $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Net current is $ ( 20-6+12-7+18 )A $ , i.e, 37A $ r=\frac{10}{100}m=\frac{1}{10}m $ $ B=\frac{{\mu_{0}}I}{2\pi r}=\frac{4\pi \times {{10}^{-7}}\times 37\times 10}{2\pi \times 1}=74\times {{10}^{-6}}T $ $ 74\mu T. $
BETA
