A current `I` flows an infinitely long wire with cross section in the form of a semi - circular ring of radius `R`. The magnitude of the magnetic induction along its axis is :
A
`(mu_(0)I)/( 2 pi^(2) R )`
B
`(mu_(0)I)/( 2 pi R )`
C
`(mu_(0)I)/( 4 pi R )`
D
`(mu_(0)I)/( pi^(2) R )`
Text Solution
Verified by Experts
The correct Answer is:
D
Current in a small element , `dl = (d theta)/(pi) I` Magnetic field due to the element `dB = (mu_(0))/( 4 pi) ( 2dI)/( R )` The component `dB cos theta`, of the field is cancelled by another opposite component. Therefore, `B_( net) =int dB sin theta = (mu_(0)I)/( 2 pi^(2) R_(0)) int_(0)^(pi) sin theta d theta = (mu_(0)I)/( pi^(2)R )`
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