A long solenoid having n = 200 turns per metre has a circular cross-section of radius `a_(1) = 1 cm`. A circular conducting loop of radius `a_(2) = 4 cm` and resistance ` R = 5 (Omega)` encircles the solenoid such that the centre of circular loop coincides with the midpoint of the axial line of the solenoid and they have the same axis as shown in Fig.

A current 't' in the solenoid results in magnetic field along its axis with magnitude `B = (mu)ni` at points well inside the solenoid on its axis. We can neglect the insignificant field outside the solenoid. This results in a magnetic flux `(phi)_(B)` through the circular loop. If the current in the winding of solenoid is changed, it will also change the magnetic field `B = (mu)_(0)ni` and hence also the magnetic flux through the circular loop. Obvisouly, it will result in an induced emf or induced electric field in the circular loop and an induced current will appear in the loop. Let current in the winding of solenoid be reduced at a rate of `75 A //sec`.
When the current in the solenoid becomes zero so that external magnetic field for the loop stops changing, current in the loop will follow a differenctial equation given by [You may use an approximation that field at all points in the area of loop is the same as at the centre

Options

A. `vecr=(hati+2hatj-hatk)+lamda(-hati+5hatj-3hatk)`
B. `vecr=hati+3hatj+2hatk+lamda(hati-5hatj+3hatk)`
C. `vecr=hati+3hatj+2hatk+lamda(hati+5hatj+3hatk)`
D. `vecr=hati+3hatj+2hatk+lamda(-hati-5hatj-3hatk)`

Answer

b

Explanation

The required line passes through the point `hati+3hatj+2hatk` and is perpendicular to the lines `vecr= (hati+2hatj-hatk)+ lamda(2hati+hatj+hatk) and vecr= (2hati+ 6hatj+hatk)+mu (hati+ 2hatj+3hatk)`, therefore, it is parallel to the vector
`vecb= (2hati+hatj+ hatk)xx (hati+2hatj+3hatk)=(hati-5hatj+3hatk)`
Hence, the equation of the required line is
`" "vecr= (hati+3hatj+2hatk)+lamda(hati-5hatj+3hatk)`