Question

You have designed and constructed a solenoid to produce a magnetic field equal in magnitude to that of the Earth (5.0 10-5 T). If your solenoid has 350 turns and is 32 cm long, determine the current you must use in order to obtain a magnetic field of the desired magnitude.

Answer 1

Answer:

$I = 0.03637 A$

Explanation:

The given data in the question is

Magnetic field : $B = 5.0 \times 10^{-5} T$

Turns : $N = 350$

Length : $L = 32 cm = 0.32 m$

So, number of turns per unit length :

$n=\frac{N}{L}$

$n=\frac{350\: turns}{0.32 m}$

$n=1093.75 \: turns \: per \: meter$

If current is I , then magnetic field is given by

$B = \mu_{0} \times n \times I$

Also,

$I = \frac{B}{\mu_{0} \times n}$

Insert the values

$I = \frac{5.0 \times 10^{-5}}{4 \pi \times 10^{-7} \times 1093.75} A$

$I = 0.03637 A$

The current will be $I = 0.03637 A$