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

Options

A. `1.5xx10^(-6)N m`
B. `1.5xx10^(-8)N m`
C. `1.5xx10^(+6)N m`
D. `1.5 xx10^(+8)N m`

Answer

B

Explanation

Here,
For small circular coil,
Number of turns ,N =10
Area ,A=1 `mm^(2)=1xx10^(-6)m^(2)`
Current,`I_(t)=(21)/(44)A`
For a long solenoid
Number of turns per metre ,`n=10^(3)` per m
Current ,`I_(2)=2.5A`
Magnetic field due to a long solenoid on its axis is
`B=mu_(0)nl_(2)`........(i)
Magnetic moment of a circular coil is
M=`NAI_(1)`..........(ii)
Torque,`vectau=vecMxxvecB`
`tau=MBsintheta=MB(becausetheta=90^(@)`(Given))
`tau=(NAI_(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