For a wire, as shown in the figure, carrying a current of 10 A, then the magnetic induction field at the point O is [take, `mu_(0)=4pi xx 10 ^(-7)H//m]`

Options

A. `2 xx 10 ^(-4) T`
B. ` 4 xx 10 ^(-4)T`
C. `10 xx 10 ^(-4) T`
D. `4pi xx 10^(-4)T`

Answer

B

Explanation

Magnetic field due to a long current carrying wire at aone end is given by `B = (mu_(0)i)/(4pid)'`
(where, I = current and d= distance from the wire) For both the wire,
`I = 10A, d =1/2 xx 10 ^(-2) m`
Magnetic field due to one wire,
`B= (mu_(0)xx10)/(4pi xx ((1)/(2)xx 10 ^(-2)))`
`B = (2xx 10 ^(-7) xx 10)/(10 ^(-2)) = 2xx 10 ^(-4)T`
Due to both wires magnetic field is in same direactin, so they will add up.
`implies` Total magnetic field, `B = 2 (2xx 10 ^(-4)) `
`B = 4xx 10 ^(-4) T`