Question

A long straight wire carrying a current of 30 A is placed in an external uniform magnetic field of induction $4\times {10}^{-4}T.$ The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point 2.0 cm away from the wire is $({\mu }_0=4\pi \times {10}^{-7}H/m)$

• A. ${10}^{-4}$
• B. $3\times {10}^{-4}$
• C. $5\times {10}^{-4}$
• D. $6\times {10}^{-4}$

Answer 1

The correct option is C $5\times {10}^{-4}$

Magnetic field induction at point P due to current carrying wire is

$B_2=\frac{{\mu }_{0}i}{2\pi r}.\frac{4\pi \times {10}^{-7}\times 30}{2\pi \times 0.02}$

$B_2=3\times {10}^{-4}T$

The direction of $B_2$ will be perpendicular to $B_1$ then, the magnitude of the resultant magnetic induction

$B=\sqrt{B_1^2+B_2^2}$

$=\sqrt{(4{)}^2+(3{)}^2\times {10}^{-4}}$

$=5\times {10}^{-4}T$