Question

What is the self - inductance of an air-core solenoid, 1m long and 0.05m in diameter, if it has 1400 turns?
5.23mH
4.84mH
3.63mH
2.42mH

17 Which of the following is NOT correct?
A changing electric field can produce a changing magnetic field
A steady magnetic field produces a steady current
A changing magnetic field can produce a changing current
A changing magnetic field can produce a steady electric field

18 A rectangular coil of dimensions 20cm by 15cm lies with its plane parallel to a magnetic field of
0.5W/m2
. The coil, carrying a current of 10A experiences a torque of 4.5Nm in the field. How many loops has the coil?
100
60
30
20

Answer 1

Answer on Question 55899, Physics, Electromagnetism

What is the self-inductance of an air-core solenoid, $1m$ long and $0.05m$ in diameter, if it has 1400 turns?

a) $5.23mH$

b) $4.84mH$

c) $3.63mH$

d) $2.42mH$

Solution:

We can find the self-inductance of an air-core solenoid from the formula:

$$L = \mu_0 \frac{N^2 A}{l} = \mu_0 \frac{N^2 \frac{\pi d^2}{4}}{l} = \mu_0 \frac{N^2 \pi d^2}{4l},$$

where, $\mu_0$ is the permeability of free space, $N$ is the number of turns, $A = \frac{\pi d^2}{4}$ is the cross-sectional area, $d$ is the diameter of the wire, $l$ is the length of solenoid.

Let's calculate the self-inductance of an air-core solenoid:

$$L = 4\pi \cdot 10^{-7} \frac{H}{m} \cdot \frac{(1400)^2 \cdot \pi \cdot (0.05m)^2}{4 \cdot 1m} = 4.84 \cdot 10^{-3} H = 4.84mH.$$

Answer:

b) $4.84mH$

17. Which of the following is not correct:

a) A changing electric field can produce a changing magnetic field.

b) A steady magnetic field produces a steady current.

c) A changing magnetic field can produce a changing current.

d) A changing magnetic field can produce a steady electric field.

Answer:

A static magnetic field relative to a wire induces a zero current.

Therefore, the false statement is b) A steady magnetic field produces a steady current.

18. A rectangular coil of dimensions $20cm$ by $15cm$ lies with its plane parallel to a magnetic field of $0.5\frac{W}{m^2}$ . The coil, carrying a current of $10A$ experiences a torque of $4.5Nm$ in the field. How many loops has the coil?

a) 100

b) 60

c) 30

d) 20

Solution:

Let us consider a rectangular loop of coil carrying a current $I$ in the presence of a uniform magnetic field $B$ directed parallel to the plane of the loop:

Figure 1.

We see in the Fig. 1a, that no magnetic forces act on sides 1 and 3 because these wires are parallel to the field. However, magnetic forces do act on sides 2 and 4 because these

sides are oriented perpendicular to the field. We can obtain the magnitude of this forces from the equation $\pmb{F}_{B} = I\pmb{L} \times \pmb{B}$ , where $\pmb{F}_{B}$ is the magnetic force, $I$ is the current in the wire, $\pmb{L}$ is a vector that points in the direction of the current $I$ and has a magnitude equal to the length $L$ of the wire, $\pmb{B}$ is the magnetic field. So, the magnitude of this forces is:

$$F _ {2} = F _ {4} = I a B$$

The direction of $F_{2}$ , the force exerted on wire 2 is out of the page in the view shown in the Fig. 1a, and that of $F_{4}$ , the force exerted on wire 4, is into the page in the same view. If we view the loop from side 3 and sight along sides 2 and 4, we see the view shown in Fig. 1b, and the two forces $F_{2}$ and $F_{4}$ are directed as shown. So, these two forces produce about point $O$ a torque and the magnitude of this torque $\tau$ is:

$$\tau = F _ {2} \frac {b}{2} + F _ {4} \frac {b}{2} = (I a B) \frac {b}{2} + (I a B) \frac {b}{2} = I a b B$$

where the moment arm about point $O$ is $\frac{b}{2}$ for each force.

Because the torque increases proportionally according to number of loops $N$ we obtain:

$$\tau = N I a b B$$

Finally, we can find the number of loops of a rectangular coil:

$$N = \frac {\tau}{I a b B} = \frac {4 . 5 N m}{1 0 A \cdot 0 . 2 m \cdot 0 . 1 5 m \cdot 0 . 5 \frac {W}{m ^ {2}}} = 3 0 l o o p s.$$

Answer:

c) 30 loops

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