KEAM 2018 Physics and Chemistry Paper

Magnetic field due to a straight current carrying conductor of finite length
B=‌

µ0

4π

‌

I

d

(sin‌θ1+sin‌θ2)....(i)
(i) Magnetic field due to conductor DA.
Here, d=‌

4

2

=2m
θ1=θ2=tan−1(‌

2√2

2

)=54.73∘
and ‌‌I=5A (given)
From Eq. (i),
B1=‌

µ0

4π

×‌

5

2

(0.816+0.816)
B1=4.08×10−7T
Similarly, magnetic field due to conductor BC
B2=4.08×10−7T
(ii) Magnetic field due to conductor AB.
Here, ‌‌d=‌

4√2

2

=2√2m
θ1=θ2=tan−1(‌

2

2√2

)=35.26∘
and I=5A (given)
From Eq. (i),
B3=‌

µ0

4π

×‌

5

2√2

(0.57+0.57)
B3=2.04×10−7T
Similarly magnetic field due to conductor CD
B4=2.04×10−7T
Hence, magnitude of induction field vector B at the intersection of the diagonals
B=B1+B2+B3+B4
B=12.24×10−7T
or B=1.2×10−6T