A long straight wire carrying a current of `25A` is placed in an external uniform magnetic field `3*0xx10^-4T` parallel to the current. Find the magnitude of the resultant magnetic field at a point `1*5cm` away from the wire.

A long straight wire carrying a current of `25A` is placed in an exter

1.	A long straight wire carrying a current of `25A` is placed in an external uniform magnetic field `30xx10^-4T` parallel to the current. Find the magnitude of the resultant magnetic field at a point `15cm` away from the wire.
Answer» Correct Answer - `448xx10^-4T` Here, `I=25A`, `r=15xx10^-2m`, `B_2=3xx10^-4T` Magnetic field due to straight current carrying wire, `B_1=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx25)/((15xx10^-2))=10/3xx10^-4T` This magnetic field will act perpendicular to external magnetic field `B_2(=3xx10^-4T)`. Therefore, the magnitude of the resultant magnetic field is `B=sqrt(B_1^2+B_2^2)=sqrt((10/3xx10^-4)^2+(3xx10^-4)^2)` `=448xx10^-4T`

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A long straight wire carrying a current of `25A` is placed in an external uniform magnetic field `3*0xx10^-4T` parallel to the current. Find the magnitude of the resultant magnetic field at a point `1*5cm` away from the wire.

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A long straight wire carrying a current of `25A` is placed in an external uniform magnetic field `30xx10^-4T` parallel to the current. Find the magnitude of the resultant magnetic field at a point `15cm` away from the wire.