Consider a solenoid of length L, N windings, and radius b (L is much longer than b). A current I is flowing through the wire. If the length of the solenoid became twice as long (2L), and all other quantities remained the same, the magnetic field inside the solenoid would Consider a solenoid of length , windings, and radius ( is much longer than ). A current is flowing through the wire. If the length of the solenoid became twice as long (2), and all other quantities remained the same, the magnetic field inside the solenoid would:. a. stay the same.
b. become twice as strong as initially.
c. become half as strong as initially.

The magnetic field inside the solenoid would be 1/2

Explanation:

The magnitude of the magnetic field (B) inside a solenoid is given by equation –  

B=μonIB=μonI

Where

I is the current  

n is the number of turn per unit  

n = N/L , where N is the number of turns(windings) and L is the length of the solenoid.

μoμo is the permeability constant

So the formula becomes

B=μo(N)IB/L

If L is doubled, magnetic flux becomes half the original value

close lines strong magnet less close weaker

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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