The torque required to hold a small circular coil of 10 turns, area `1mm^(2)` and carrying a current of `((21)/(44))A` in the middle of a long solenoid of `10 ^(3)"turns"//"m"` carrying a current of 2.5A, with its axis perpendicular to the axis of the solenoid is

Options

A. `1.5xx10^(-6)Nm`
B. `1.5xx10^(-8)Nm`
C. `1.5xx10^(6)Nm`
D. `1.5xx10^(8)Nm`

Answer

B

Explanation

Here,
For small circular coil,
Number of turns, `N` = 10, Area , `A`=1 `"mm"^(2)` = `1xx10^(-6)" m"^(2)`
Current, ` I_(1)=(21)/(44)A `
For a long solenoid,
Number of turns per metre, `n=10^(3)` per m
Current, `I_(2)=2.5` A
Magnetic field due to a long solenoid on its axis is
`B=mu_(0)nI_(2)" "...(i)`
Magnetic moment of a circular coil is
` M= NAI_(1)" "...(ii)`
` vec(tau)=vec(M) xx vec(B) `
` tau=MBsintheta=MB" "(because theta=90^(@)("Given"))`
`tau=(NA I_(1))(mu_(0)nI_(2))" "("Using"(i)and (ii)) `
` tau=10xx1xx10^(-6) xx(21)/(44 )xx4xx(22)/(7)xx10^(-7)xx10^(3)xx2.5`
`=1.5xx10^(-8)" N m"`