KEAM 2021 Physics and Chemistry Solved Paper

Given, length of conductor,
l=20cm=0.2m
Current, i=5A,  ‌‌θ=30∘
Magnetic field,  ‌‌‌B=0.5T
Force acting on the conductor in the external magnetic field is given as
F=il×B
In terms of magnitude,
F=ilB‌sin‌θ
Substituting the given values, we get
F=5×0.2×0.5×sin‌30∘
=5×0.2×0.5×‌

1

2

=0.25N