A winding wire which is used to frame a solenoid can bear a maximum $10\, A$ current. If length of solenoid is $80\,cm$ and it's cross sectional radius is $3\, cm$ then required length of winding wire is $(B = 0.2\,T)$
Medium
Download our app for free and get started
(c) $B = \frac{{{\mu _0}Ni}}{l}$ where $N =$ Total number of turns, $l =$ length of the solenoid
$ \Rightarrow 0.2 = \frac{{4\pi \times {{10}^{ - 7}} \times N \times 10}}{{0.8}} \Rightarrow N = \frac{{4 \times {{10}^4}}}{\pi }$
Since $N$ turns are made from the winding wire so length of the wire $(L)$ $ = 2\pi r \times N$ [ $2\pi r = $ length of each turns]
$ \Rightarrow L = 2\pi \times 3 \times {10^{ - 2}} \times \frac{{4 \times {{10}^4}}}{\pi } = 2.4 \times {10^3}\,m$
$ \Rightarrow 0.2 = \frac{{4\pi \times {{10}^{ - 7}} \times N \times 10}}{{0.8}} \Rightarrow N = \frac{{4 \times {{10}^4}}}{\pi }$
Since $N$ turns are made from the winding wire so length of the wire $(L)$ $ = 2\pi r \times N$ [ $2\pi r = $ length of each turns]
$ \Rightarrow L = 2\pi \times 3 \times {10^{ - 2}} \times \frac{{4 \times {{10}^4}}}{\pi } = 2.4 \times {10^3}\,m$
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 charge moving with velocity $v$ in $X$-direction is subjected to a field of magnetic induction in the negative $X$-direction. As a result, the charge willView Solution
- 2A conducting wire bent in the form of a parabola $y^2 = 2x$ carries a current $i = 2 A$ as shown in figure. This wire is placed in a uniform magnetic field $\vec B = - 4\,\hat k$ $Tesla$. The magnetic force on the wire is (in newton)View Solution
- 3A straight wire $AB$ of mass $40\,g$ and length $50\,cm$ is suspended by a pair of flexible leads in uniform magnetic field of magnitude $0.40\,T$ as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ...........$A$. (Take $g=10\,ms ^{-2}$ ).View Solution
- 4The magnetic field at the origin due to a current element $i\,\overrightarrow {dl} $ placed at position $\vec r$ isView Solution
$(i)\,\,\left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{d\vec l\, \times \,\vec r}}{{{r^3}}}} \right)$
$(ii)\,\, - \left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{d\vec l\, \times \,\vec r}}{{{r^3}}}} \right)$
$(iii)\,\left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{\,\vec r \times d\vec l}}{{{r^3}}}} \right)$
$(iv)\, - \left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{\,\vec r \times d\vec l}}{{{r^3}}}} \right)$
- 5A circular coil of wire carries a current. $PQ$ is a part of a very long wire carrying a current and passing close to the circular coil. If the directions of currents are those shown in figure, what is the direction of force acting on $PQ$ ?View Solution
- 6One of the two identical conducing wires of length $L$ is bent in the form of a circular loop and the other one into a circular coil of $N$ identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop $(B_L)$ to that at the centre of the coil $(B_C),$; $.\,\frac {B_L}{B_C}$ will beView Solution
- 7A particle of mass $'m'$ and carrying a charge $'q'$ enters with a velocity $'v'$ perpendicular to a uniform magnetic field. The time period of rotation of the particleView Solution
- 8The magnetic field at the centre of a circular coil of radius $r$ is $\pi $ times that due to a long straight wire at a distance $r$ from it, for equal currents. Figure here shows three cases : in all cases the circular part has radius $r$ and straight ones are infinitely long. For same current the $B$ field at the centre $P$ in cases $1$, $2$, $ 3$ have the ratioView Solution
- 9The resistance of a galvanometer is $50\,\Omega $ and it requires $2\,\mu A$ per two division deflection. The value of the shunt required in order to convert this galvanometer into ammeter of range $5\,A$ is (The number of divisions on the galvanometer scale on one side is $30$)View Solution
- 10A magnetised wire of moment $M$ is bent into an arc of a circle subtending an angle of $60^o$ at the centre; then the new magnetic moment isView Solution