Moving Charges and Magnetism

Physics

NCERT

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?

Solution :

Number of turns on the circular coil, $n = 100$$\\$ Radius of each turn,$ r = 8.0 cm =0.08 m$$\\$ Current flowing in the coil, $I = 0.4 A$$\\$ Magnitude of the magnetic field at the centre of the coil is given by the relation,$\\$ $|B|=\dfrac{\mu_0}{4\pi}\dfrac{2\pi nl}{r}$$\\$ $\mu_0=$Permeability of free space$\\$ $=4\pi * 10^{-7}TmA^{-1}\\ |B|=\dfrac{4\pi *10^{-7}}{4 \pi} * \dfrac{2 \pi * 100 * 0.4}{0.08}\\ =3.14*10^{-4}T$$\\$ Hence, the magnitude of the magnetic field is $3.14*10^{-4} T .$

2   A long straight wire carries a current of $35 A.$ What is the magnitude of the field $B$ at a point $20 cm$ from the wire?

Solution :

Current in the wire,$ I =35 A$$\\$ Distance of a point from the wire, $r = 20 cm = 0.2 m$$\\$ Magnitude of the magnetic field at this point is given as:$\\$ $B=\dfrac{\mu_0}{4\pi}\dfrac{2l}{r}$$\\$ Where,$\\$ $\mu_0=$Permeability of free space$=4\pi*10^{-7}TmA^{-1}$$\\$ $B=\dfrac{4\pi*10^{-7}*2*35}{4\pi *0.2}\\ =3.5*10^{-5}T$$\\$ Hence, the magnitude of the magnetic field at a point $20 cm$ from is $3.5 *10^{-5} T .$

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.

Solution :

Current in the wire, $I = 50 A$$\\$ A point is $2.5 m$ away from the East of the wire.$\\$ $\therefore $ Magnitude of the distance of the point from the wire,$ r =2.5 m$ .$\\$ Magnitude of the magnetic field at that point is given by the relation,$\\$ $B=\dfrac{\mu_0 2l}{4\pi r}$$\\$ Where,$\\$ $\mu_0=$Permeability of free space$=4\pi*10^{-7}TmA^{-1}$$\\$ $B=\dfrac{4\pi*10^{-7}*2*50}{4\pi*2.5}\\ =4*10^{-6}T$$\\$ 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 outward. Hence, according to the Maxwell’s right and thumb rule, the direction of the magnetic field at the given point is vertically upward.

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?

Solution :

Current in the power line, $I = 90 A$$\\$ Point is located below the power line at distance, $r = 1.5 m$$\\$ Hence, magnetic field at that point is given by the relation,$\\$ $B=\dfrac{\mu_02l}{4 \pi r}$$\\$ Where,$\\$ $\mu_0=$Permeability of free space$=4\pi*10^{-4}TmA^{-1}$$\\$ $B=\dfrac{4 \pi * 10^{-7}*2 * 90}{4 \pi *1.5}\\ =1.2*10^{-5}T$$\\$ 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.

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 $30^o$ with the direction of a uniform magnetic field of $0.15 T$?

Solution :

Current in the wire,$ I = 8 A$$\\$ Magnitude of the uniform magnetic field, $B = 0.15 T$$\\$ Angle between the wire and magnetic field, $\theta=30^o$$\\$ Magnetic force per unit length on the wire is given as:$\\$ $f=BI \sin \theta \\ =0.15* 8 * 1* \sin 30^o\\ =0.6 N m^{-1}$$\\$ Hence, the magnetic force per unit length on the wire is $0.6 N m^{-1} $.