Q113 / 150 · MHT CET 2019 (Shift 1) · Physics

Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are l1=+10A,l2=-13A,l3=+10A,l4=+7A,l5=-12A and l6=+18A . The magnetic induction at a perpendicular distance of 10 cm from the cable is μ0=4π×10-7WbA-m

  • Correct answer
    A
    40μT
  • B
    37.5μT
  • C
    30μT
  • D
    35μT

Solution

Net current due to all wires,
inet=i1+i2+i3+i4+i5+i6
inet=10+13+10+7-12+18=20A
We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance r from it is given by
B=μ0i2πr=μ0inet2πr
where, i = current in wire
and r = perpendicular distance.
=4π×10-7×202π×10×10-2=4×10-5
B=40μT

Hence, the correct answer is option: 'A'.

Relevant Questions