AP EAPCET 07-Jul-2022 Shift 2 Paper

Given arrangement is shown below :

Distance of conductor I from origin, is
d=√x2+y2=√1+1
⇒d=√2‌cm=√2×10−2m
Magnetic field due to wire Iat origin is
B1=

µ0I1

2πd

or B1=B0=

µ0

2πd

...[∴I1=I2=1A]
Now, in vector form, B1=

µ0

2πd

(

1

√2

^

i

−

1

√2

^

j

)
=

µ0

2π×√2×10−2

(

1

√2

^

i

−

1

√2

^

j

)
And field of conductor II is
B2=

µ0I2

2πy

(−

^

k

)
=

µ0

2π×10−2

(−

^

k

) ...[∴y=1‌cm=1×10−2m]
Hence, magnetic field due to both conductors,
B=B1+B2
=

µ0

2π

(

1

√2

^

i

−

1

√2

^

j

)×

1

√2×10−2

+

µ0

2π

(−

^

k

)×

1

10−2

⇒B=

µ0

2π×10−2

{

^ i

2

−

^ j

2

−

^

k

}
Hence, ratio

B

B0

is

B

B0

=(

µ0 2π×10−2 ( ^ i 2 − ^ j 2 − ^ k )

µ0 2π×√2×10−2

)=(

^ i

2

−

^ j

2

−

^

k

)√2

^ i

√2

−

^ j

√2

−√2

^

k

B

B0

=(

1

√2

,−

1

√2

,−√2)