A 20-cm long solenoid consists of 100 turns of a coil of radius r = 3.0 cm. A current of I_0 in the coiled wire produces a magnetic field of B_0 in the middle of the solenoid.
If we wish to produce a field that is 4-times larger, which of the following changes could work?

a) quadruple the number of turns in the 20-cm long solenoid.
b) half the length of the 100-turn solenoid but maintain 100 turns across that length.
c) quadruple the length of the 100-turn solenoid, but also double the radius of the coil.
d) half the length of the solenoid, and double the number of turns to 200 total turns.
e) double the current in the wire, and double the number of turns in the 20-cm long solenoid.
f) cut the radius of the coil to one-half.

Explanation:

length of the solenoid, l = 20 cm = 0.2 m

radius, r = 3 cm

current in the coil = Io

Magnetic field = Bo

The magnetic field due to the current carrying solenoid is

..... (1)

where, n is the number of turns per unit length.

so, increasing the number of turns by four times in 20 cm length solenoid, the magnetic field is 4 times. option (a)

And, by taking half the length of the solenoid, and the double of number of turns, the magnetic field due to the solenoid becomes 4 times. option (d)

a.   mass number from atomic number

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west