A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the center of the coil is
A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the center of the coil is
A. `2.01 xx 10^(-3)T`
B. `5.64 xx 10^(-3)T`
C. `2.64 xx 10^(-3)T`
D. `5.64 xx 10^(-3)T`
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Correct Answer - A
As `B=(mu_(0)NI)/(2R)," Here N "= 100, I = 3.2 A,`
`R = 10 cm = 10xx10^(-2)m`
`therefore" "b=(4pixx10^(7)xx100xx3.2)/(2xx0.1)=2.01xx10^(-3)T`
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