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Chapter 19: Problem 74

A solenoid has 4850 turns per meter and radius \(3.3 \mathrm{cm}\) The magnetic field inside has magnitude 0.24 T. What is the current in the solenoid?

Short Answer

Expert verified
Answer: The current flowing through the solenoid is approximately 3.11 A.

Step by step solution

01

Write down the given values and the formula for the magnetic field inside a solenoid.

The given values are number of turns per meter (\(n = 4850 \, turns/m\)), radius (\(r = 3.3 \, cm\)), and the magnetic field inside the solenoid (\(B = 0.24 \, T\)). The permeability of free space (\(\mu_0 = 4\pi \times 10^{-7} \, Tm/A\)) is a constant value. We will use Ampère's law: $$ B = \mu_0 \cdot n \cdot I $$
02

Solve the formula for current I.

We want to find the current I in the solenoid. Therefore, rearrange the formula to get I by itself: $$ I = \frac{B}{\mu_0 \cdot n} $$
03

Plug in the given values and solve for I.

Now, we can substitute the given values into the formula and calculate the current I: $$ I = \frac{0.24 T}{(4\pi \times 10^{-7} \, Tm/A)(4850 \, turns/m)} $$ After plugging in the numbers and doing the calculations, we get: $$ I \approx 3.11 A $$ So the current in the solenoid is approximately 3.11 A.

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Most popular questions from this chapter

Two parallel wires in a horizontal plane carry currents \(I_{1}\) and \(I_{2}\) to the right. The wires each have length \(L\) and are separated by a distance \(d\). (a) What are the magnitude and direction of the ficld due to wire 1 at the location of wire \(2 ?\) (b) What are the magnitude and direction of the magnetic force on wire 2 due to this field? (c) What are the magnitude and direction of the field due to wire 2 at the location of wire \(1 ?\) (d) What are the magnitude and direction of the magnetic force on wire 1 due to this field? (e) Do parallel currents in the same direction attract or repel? (f) What about parallel currents in opposite directions?
In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=2.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) to the east and a uniform magnetic field \(\overline{\mathbf{B}}=0.0050 \mathrm{T}\) to the west. (a) What is the electromagnetic force on an electron moving north at \(1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?(\mathrm{b})\) With the electric and magnetic fields as specified, is there some velocity such that the net electromagnetic force on the electron would be zero? If so, give the magnitude and direction of that velocity. If not, explain briefly why not.

An electromagnetic rail gun can fire a projectile using a magnetic field and an electric current. Consider two conducting rails that are \(0.500 \mathrm{m}\) apart with a \(50.0-\mathrm{g}\) conducting rod connecting the two rails as in the figure with Problem \(46 .\) A magnetic field of magnitude \(0.750 \mathrm{T}\) is directed perpendicular to the plane of the rails and rod. A current of \(2.00 \mathrm{A}\) passes through the rod. (a) What direction is the force on the rod? (b) If there is no friction between the rails and the rod, how fast is the rod moving after it has traveled \(8.00 \mathrm{m}\) down the rails?
The intrinsic magnetic dipole moment of the electron has magnitude $9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ In other words, the electron acts as though it were a tiny current loop with $N I A=9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ What is the maximum torque on an electron due to its intrinsic dipole moment in a \(1.0-T\) magnetic field?
The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). What must be the minimum radius of the dees if the maximum proton speed desired is $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?$
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