An ideal solenoid of radius a has n turns per unit length and current I.The magnetic flux $\phi$ _B through any area completely inside the solenoid,centered on the solenoid axis but at a 45$\degree$angle to the axis,so that it touches the inside of the solenoid,as shown below,is A) $\mu _ { 0 } \frac { \pi a ^ { 2 } } { 4 } n I$ . B) $\mu _ { 0 } \frac { \pi a ^ { 2 } } { 2 } n I$ . C)$\mu$₀ $\pi$a²nI. D)2$\mu$₀ $\pi$a²nI. E)0.

Question

Quizplus · Accepted Answer

The flux through any surface that is completely inside the solenoid and parallel to its axis is given by 
Φ = μ0nIA,
where μ0 is the permeability of free space, n is the number of turns per unit length, I is the current, and A is the area of the surface. 
Here, the surface is at an angle of 45° to the axis, but it still touches the inside of the solenoid. So, we can consider the component of A that is parallel to the axis of the solenoid, which is A cos 45° = A/√2. 
Therefore, 
Φ = μ0nI(A/√2) = (μ0nIA/√2). 
Substituting the given values, we get 
Φ = (4π×10-7)(nI)(πa2/2√2) = (π2a2nI/2√2). 
Hence, the answer is (C).

An Ideal Solenoid of Radius a Has N Turns Per ϕ\phiϕ

An Ideal Solenoid of Radius a Has N Turns Per $\phi$