Question : A 1 cm segment of a wire lying along x-axis carries current of 0.5 A along +x direction. A magnetic field B = (0.4 mT) j + (0.6 mT) k is switched on, in the region. The force acting on the segment is
(A) (2j+3k) mN
(B) (-3j+2k) µM
(C) (6j+4k) mN
(D) (-4j+6k) µN
Doubt by Niyati
Solution :
l=(1 cm) i
l=(10-2 m) i
I = 0.5 A
B=(0.4 mT) j + (0.6 mT) k
B=(0.4×10-3 T) j + (0.6×10-3 T) k
We know,
F=I(l×B)
F=0.5{10-2 i×[(0.4×10-3) j+(0.6×10-3) k]}
F=0.5{(10-2)(0.4×10-3)k+(10-2)(0.6×10-3)(-j)}
F=0.5{(0.4×10-5)k-(0.6×10-5)j}
F=(0.20×10-5)k-(0.30×10-5)j}
F=(0.20×10-5)k-(0.30×10-5)j}
F=(2.0×10-6)k-(3.0×10-6)j}
F=(-3j+2k)×10-6
F=(-3j+2k)µN
Hence, (B) (-3j+2k) µM, would be the correct option.
