A coil of self inductance 20 mH, having 50 turns, carries a current of 300 mA. If the area of the coil is `2cm^(2)`, the magnetic induction at the cen
A coil of self inductance 20 mH, having 50 turns, carries a current of 300 mA. If the area of the coil is `2cm^(2)`, the magnetic induction at the centre of the coil is
A. `7.5xx10^(-3)T`
B. `7.5xx10^(-2)T`
C. `6xx10^(-1)T`
D. `0.5T`
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Correct Answer - C
`B=(mu_(0)nI)/(2a)`
Now `A=pia^(2)" "therefore a=sqrt((A)/(pi))`
`=(4xx10^(-7)xx50xx300xx10^(-3))/(2xxsqrt((2xx10^(-4))/(pi)))`
`=0.6T`
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