Question 1

Q. 4.1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A . What is the magnitude of the magnetic field B at the centre of the coil?

Accepted Answer

Number of turns on the circular coil, $\mathrm{n}=100$

Radius of each turn, $\mathrm{r}=8.0 \mathrm{~cm}=0.08 \mathrm{~m}$

Current flowing in the coil, $\mathrm{I}=0.4 \mathrm{~A}$

Magnitude of the magnetic field at the centre of the coil is given by the relation,

$$

|\vec{B}|=\frac{\mu_{0}}{4 \pi} \frac{2 \pi n l}{r}

$$

Where,

$\mu_{0}=$ Permeability of free space

$$

=4 \pi 	imes 10^{-7} \mathrm{Tm} \mathrm{~A}^{-1}

$$

So,

$$

\begin{aligned}

|\vec{B}| & =\frac{4 \pi 	imes 10^{-7}}{4 \pi} 	imes \frac{2 \pi 	imes 100 	imes 0.4}{r} \

& =3.14 	imes 10^{-4} T

\end{aligned}

$$

Hence, the magnitude of the magnetic field is $3.14 	imes 10^{-4} \mathrm{~T}$.

Question 2

Q. 4.2. Along straight wire carries a current of 35 A . What is the magnitude of the field B at a point 20 cm from the wire?

Accepted Answer

Current in the wire, $\mathrm{I}=35 \mathrm{~A}$

Distance of a point from the wire, $\mathrm{r}=20 \mathrm{~cm}=0.2 \mathrm{~m}$

Magnitude of the magnetic field at this point is given as:

$$

|\vec{B}|=\frac{\mu_{0}}{4 \pi} \frac{2 l}{r}

$$

Where,

$\mu_{0}=$ Permeability of free space

$$

=4 \pi 	imes 10^{-7} \mathrm{Tm} \mathrm{~A}^{-1}

$$

\begin{aligned}

|\vec{B}| & =\frac{4 \pi 	imes 10^{-7}}{4 \pi} 	imes \frac{2 	imes 35}{0.2} \

& =3.5 	imes 10^{-5} \mathrm{~T}

\end{aligned}

$$

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is $3.5 	imes 10^{-5} \mathrm{~T}$.

Question 3

Q. 4.3. A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of $B$ at a point 2.5 m east of the wire.

Accepted Answer

Current in the wire, $\mathrm{I}=50 \mathrm{~A}$

A point is 2.5 m away from the East of the wire.

$	herefore$ Magnitude of the distance of the point from the wire, $\mathrm{r}=2.5 \mathrm{~m}$.

Magnitude of the magnetic field at that point is given by the relation,

$$

|\vec{B}|=\frac{\mu_{0}}{4 \pi} \frac{2 l}{r}

$$

Where,

$\mu_{0}=$ Permeability of free space

$$

=4 \pi 	imes 10^{-7} \mathrm{Tm} \mathrm{~A}^{-1}

$$

|\vec{B}|=\frac{4 \pi 	imes 10^{-7}}{4 \pi} 	imes \frac{2 	imes 50}{2.5}=4 	imes 10^{-6} T

$$

The point is located normal to the wire length at a distance of 2.5 m . The direction of the current in the wire is vertically downward. Hence, according to the Maxwell's right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

Question 4

Q. 4.4. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Accepted Answer

Current in the power line, $\mathrm{I}=90 \mathrm{~A}$

Point is located below the power line at distance, $\mathrm{r}=1.5 \mathrm{~m}$

Hence, magnetic field at that point is given by the relation,

$$

|\vec{B}|=\frac{\mu_{0}}{4 \pi} \frac{2 l}{r}

$$

Where,

$$

\begin{aligned}

\mu_{0} & =	ext { Permeability of free space } \

& =4 \pi 	imes 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}

\end{aligned}

$$

\begin{aligned}

|\vec{B}| & =\frac{4 \pi 	imes 10^{-7}}{4 \pi} 	imes \frac{2 	imes 90}{1.5} \

& =1.2 	imes 10^{-5} \mathrm{~T}

\end{aligned}

$$

The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell's right hand thumb rule, the direction of the magnetic field is towards the South.

Question 5

Q. 4.5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of $30^{\circ}$ with the direction of a uniform magnetic field of 0.15 T ?

Accepted Answer

Current in the wire, $\mathrm{I}=8 \mathrm{~A}$

Magnitude of the uniform magnetic field, $\mathrm{B}=0.15 \mathrm{~T}$

Angle between the wire and magnetic field, $	heta=30^{\circ}$.

Magnetic force per unit length on the wire is given as: $F=B I \sin 	heta$

$=0.15 	imes 8 	imes 1 	imes \sin 30^{\circ}$

$=0.6 \mathrm{~N} \mathrm{~m}^{-1}$

Hence, the magnetic force per unit length on the wire is $0.6 \mathrm{~N} \mathrm{~m}^{-1}$.

Question 6

Q. 4.6. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T . What is the magnetic force on the wire?

Accepted Answer

Length of the wire, $\mathrm{l}=3 \mathrm{~cm}=0.03 \mathrm{~m}$

Current flowing in the wire, $\mathrm{I}=10 \mathrm{~A}$

Magnetic field, $\mathrm{B}=0.27 \mathrm{~T}$

Angle between the current and magnetic field, $	heta=90^{\circ}$

Magnetic force exerted on the wire is given as:

$F=$ BIlsin $	heta$

$=0.27 	imes 10 	imes 0.03 \sin 90^{\circ}$

$=8.1 	imes 10^{-2} \mathrm{~N}$

Hence, the magnetic force on the wire is $8.1 	imes 10^{-2} \mathrm{~N}$. The direction of the force can be obtained from Fleming's left hand rule.

Question 7

Q. 4.7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm . Estimate the force on a 10 cm section of wire A .

Accepted Answer

Current flowing in wire $\mathrm{A}, \mathrm{I}_{\mathrm{A}}=8.0 \mathrm{~A}$

Current flowing in wire $\mathrm{B}, \mathrm{I}_{\mathrm{B}}=5.0 \mathrm{~A}$

Distance between the two wires, $\mathrm{r}=4.0 \mathrm{~cm}=0.04 \mathrm{~m}$

Length of a section of wire $\mathrm{A}, \mathrm{L}=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Force exerted on length L due to the magnetic field is given as:

$$

F=\frac{\mu_{0} I_{A} I_{B} L}{2 \pi r}

$$

Where, $\mu_{0}=$ Permeability of free space $=4 \pi 	imes 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

$$

F=\frac{4 \pi 	imes 10^{-7} 	imes 8 	imes 5 	imes 0.1}{2 \pi 	imes 0.04}=2 	imes 10^{-5} N

$$

The magnitude of force is $2 	imes 10^{-5} \mathrm{~N}$. This is an attractive force normal to A towards $B$ because the direction of the currents in the wires is the same.

Question 8

Q. 4.8. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm . If the current carried is 8.0 A , estimate the magnitude of $B$ inside the solenoid near its centre.

Accepted Answer

Length of the solenoid, $\mathrm{l}=80 \mathrm{~cm}=0.8 \mathrm{~m}$

There are five layers of windings of 400 turns each on the solenoid.

$	herefore$ Total number of turns on the solenoid, $\mathrm{N}=5 	imes 400=2000$

Diameter of the solenoid, $\mathrm{D}=1.8 \mathrm{~cm}=0.018 \mathrm{~m}$

Current carried by the solenoid, $\mathrm{I}=8.0 \mathrm{~A}$

Magnitude of the magnetic field inside the solenoid near its centre is given by the relation, $B=\frac{\mu_{0} N I}{l}$

Where, $\mu_{0}=$ Permeability of free space $=4 \pi 	imes 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

$$

\begin{aligned}

B & =\frac{4 \pi 	imes 10^{-7} 	imes 2000 	imes 8}{0.8} \

& =2.5 	imes 10^{-2} \mathrm{~T}

\end{aligned}

$$

Hence, the magnitude of the magnetic field inside the solenoid near its centre is $2.512 	imes 10^{-2} \mathrm{~T}$.

4. Moving Charges and Magnetism

Exercises - 1