Question

Figure shows a rectangular frame situated in a constant magnetic field. A wire BC of length 1 m is moved out with velocity 4 m/s. Magnetic field strength is 0.15 T. Force acting on the wire BC is:

पर्याय

• 18 N

• 1.8 N

• 0.18 N

• 0.018 N

Answer 1

0.018 N

Explanation:

For the e.m.f. induced in the wire (e) = Blv

= 0.15 × 1 × 4

= 0.6 V

Hence, the current flowing through the wire,

I = `e/R`

= `0.6/5`

= 0.12 A

∴ Force acting on the wire (F) = IlB

= 0.12 × 1 × 0.15

= 0.018 N