A thin wire of length $l$ is carrying a constant current. The wire is bent to form a circular coil. If radius of the coil, thus formed, is equal to $R$ and number of turns in it is equal to $n$, then which of the following graphs represent $(s)$ variation of magnetic field induction $(b)$ at centre of the coil
Easy
Download our app for free and get started
$(b, c)$ Since length of the wire is equal to $l$, therefore, $2\pi Rn = l$ or $n = \frac{l}{{2\pi R}}$.
Magnetic induction at centre of a circular coil is given by $B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni}}{R} = \frac{{{\mu _0}l\,i}}{{4\pi {R^2}}}$ $==>$ $B \propto \frac{1}{{{R^2}}}$
It means, when $R \to 0,\;B \to \infty $ and $R \to \infty ,\;B \to 0,$
Hence $(b)$ is correct and $(d)$ is wrong.
Substituting $R = \frac{l}{{2\pi n}}$ in $B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni}}{R}$
$B \propto {n^2}$. It means graph between $B$ and $n$ will be a parabola having increasing slope and passing through origin. Hence $(c)$ is correct and $(a)$ is wrong.
Magnetic induction at centre of a circular coil is given by $B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni}}{R} = \frac{{{\mu _0}l\,i}}{{4\pi {R^2}}}$ $==>$ $B \propto \frac{1}{{{R^2}}}$
It means, when $R \to 0,\;B \to \infty $ and $R \to \infty ,\;B \to 0,$
Hence $(b)$ is correct and $(d)$ is wrong.
Substituting $R = \frac{l}{{2\pi n}}$ in $B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni}}{R}$
$B \propto {n^2}$. It means graph between $B$ and $n$ will be a parabola having increasing slope and passing through origin. Hence $(c)$ is correct and $(a)$ is wrong.
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $i$ passes through the coil, the magnetic field at the centre isView Solution
- 2A circular loop of radius $0.0157\,m$ carries a current of $2.0\, amp$. The magnetic field at the centre of the loop is$({\mu _0} = 4\pi \times {10^{ - 7}}\,weber/amp - m)$View Solution
- 3View SolutionIn a moving coil galvanometer, to make the field radial
- 4A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\overrightarrow F$ the net force on the remaining three arms of the loop isView Solution
- 5In the following figure a wire bent in the form of a regular polygon of $n$ sides is inscribed in a circle of radius $a$. Net magnetic field at centre will beView Solution
- 6A proton of mass $1.67 \times {10^{ - 27}}\,kg$ and charge $1.6 \times {10^{ - 19}}\,C$ is projected with a speed of $2 \times {10^6}\,m/s$ at an angle of $60^\circ $ to the $X - $ axis. If a uniform magnetic field of $0.104$ $Tesla$ is applied along $Y - $ axis, the path of proton isView Solution
- 7In an ammeter $0.2\%$ of main current passes through the galvanometer. If resistance of galvanometer is $G,$ the resistance of ammeter will beView Solution
- 8View SolutionOne Tesla is equal to
- 9View SolutionThe electron in the beam of a television tube move horizontally from south to north. The vertical component of the earth's magnetic field points down. The electron is deflected towards
- 10The current of $1\,A$ is passed through a hexagonal conducting wire of side $1\,m$ . The magnetic induction at its centre $O$ in $Wb/m^2$ will beView Solution
