Q. A wire along $x$-axis carries a current 3.5 A. Find the force in newton on a 1 cm section of the wire exerted by a magnetic field $\overrightarrow{B} = (0.74\ \widehat{j} + 0.36\ \widehat{k})$ T.
A) $\left( 1.26\ \widehat{k} - 2.59\ \widehat{j} \right)10^{- 2}$ N
B) $\left( - 1.26\ \widehat{k} + 2.59\ \widehat{j} \right){\times 10}^{- 2}$ N
C) $\left( - 2.59\ \widehat{k} + 1.26\ \widehat{j} \right){\times 10}^{- 2}$ N
D) $\left( 2.59\ \ \widehat{k} - 1.26\ \widehat{j} \right){\times 10}^{- 2}$ N
Explanation
(d) $\ \overrightarrow{F} = i\lbrack\overrightarrow{I} \times \ \overrightarrow{B}\rbrack$
$= 3.5\ \lbrack 10^{- 2}\widehat{i}$ $\times \left( 0.74\ \widehat{j} - \ 0.36\ \widehat{k} \right)$
$= \left( 2.59\widehat{k} - 1.25\ \widehat{j} \right) \times 10^{- 2}$ N.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.