Two concentric coils each of radius equal to `2π` cm are placed right angles to each other. If 3 A and 4 A are the currents flowing through the two coils respectively. The magnetic induction ( in `Wb m^(-2)` ) at the center of the coils will be

Options

A. `10^(-5)`
B. `7xx10^(-5)`
C. `12xx10^(-5)`
D. `5xx10^(-5)`

Answer

D

Explanation

The situation is shown in the figure .

For coil 1 ,
Radius `R_(1) = 2pi cm = 2pi xx10^(-2)` m
Current `I_(1) = 3A`
Current , `I_(2) = 4A`
The magnetic induction at the centre O due to the current`I_(1)` in coil 1 is
`B_(1) = (mu_(0)I_(1))/(2R_(1))`
It acts in vertically upwards direction .
The magnetic induction at the centre O due the current `I_(2)` in coil 2 is
`B_(2) = (mu_(0)I_(2))/(2R_(2))`
It acts in the horizontal direction .
As `B_(1) and B_(2)` are perpendicular to each othr , so the net magnetic field at the centre O is
`B = sqrt(B_(1)^(2)+B_(2)^(2))`
`sqrt(((mu_(0)I_(1))/(2R_(1)))^(2)+((mu_(0)I_(2))/(2R_(2)))^(2))`
Sunstituting the given values , we get
`B = sqrt(((mu_(0)(3A))/ (2(2pixx10^(-2)m)))^(2)+((mu_(0)(5A))/(2(2pixx10^(-2)m)))^(2))`
` = ((4pi xx10^(7) " Wb m"^(-1)A^(-1))(5A))/((4pixx10^(-2)m))`
` = 5xx10^(-5) " Wb m "^(-2)`