A long straight wire carrying a current of `20A` is placed in an external uniform magnetic field of `3xx10^-4T` parallel to the current. Find the magnitude of resultant field at a point `2*0cm` away from the wire.

A long straight wire carrying a current of `20A` is placed in an exter

1.	A long straight wire carrying a current of `20A` is placed in an external uniform magnetic field of `3xx10^-4T` parallel to the current. Find the magnitude of resultant field at a point `2*0cm` away from the wire.
Answer» Here, `I=20A`, `B_2=3xx10^-4T`, `r=20xx10^-2m` Magnetic field due to a straight wire carrying current is `B_1=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx20)/(20xx10^-2)=2xx10^-4T` This magnetic field will act perpendicular to magnetic field `B_2(=3xx10^-4T)`. Therefore, the magnitude of the resultant magnetic field `B=sqrt(B_1^2+B_2^2)=sqrt((2xx10^-4)^2+(3xx10^-4)^2)` `=3*6xx10^-4T`

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A long straight wire carrying a current of `20A` is placed in an external uniform magnetic field of `3xx10^-4T` parallel to the current. Find the magnitude of resultant field at a point `2*0cm` away from the wire.

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