A coil having $N$ $turns$ carry a current $I$ as shown in the figure. The magnetic field intensity at point $P$ is
Medium
Download our app for free and get started
(a) From the formula, magnetic field intensity at point $P$, shown in given figure while carrying current $'i'$ and having $'N'$ turns is given by Magneti field at $P$ due to entire circular loop is
$\Rightarrow \frac{\mu_{0} i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
And hence for $N$ turns is,
$\Rightarrow \frac{\mu_{0} \mathrm{NiR}^{2}}{2\left(\mathrm{R}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$
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
- 1The ammeter has range $1\, ampere$ without shunt. the range can be varied by using different shunt resistances. The graph between shunt resistance and range will have the natureView Solution
- 2A wire length $2\,m$ carrying a current of $1\,A$ is bent to form a circle. The magnetic moment of the coil isView Solution
- 3A $2\, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5\, tesla$. The force on the proton isView Solution
- 4A cylindrical conductor of radius $'R'$ carries a current $'i'$. The value of magnetic field at a point which is $R/4$ distance inside from the surface is $10\, T$. Find the value of magnetic field at point which is $4\,R$ distance outside from the surfaceView Solution
- 5A proton and an $\alpha$ -particle, having kinetic energies $K _{ p }$ and $K _{\alpha},$ respectively, enter into $a$ magnetic field at right angles.View Solution
The ratio of the radii of trajectory of proton to that of $\alpha$ -particle is $2: 1 .$ The ratio of $K _{ p }: K _{\alpha}$ is :
- 6A galvanometer with its coil resistance $25\,\Omega $ requires a current of $1\,mA$ for its full deflection. In order to construct an ammeter to read up to a current of $2\,A,$ the approximate value of the shunt resistance should beView Solution
- 7Consider the circular loop having current $i$ and with central point $O$. The magnetic field at the central point $O$ isView Solution
- 8Given below are two statements$:$View Solution
Statement $I:$ Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element (IdI) of a current carrying conductor only.
Statement $II :$ Biot-Savart's law is analogous to Coulomb's inverse square law of charge $q$, with the former being related to the field produced by a scalar source, Idl while the latter being produced by a vector source, $q$. In light of above statements choose the most appropriate answer from the options given below:
- 9View SolutionA charged particle enters a uniform magnetic field perpendicular to it. The magnetic field
- 10If $n$ represents the actual number of deflections in a converted galvanometer of resistance $G$ and shunt resistance $S$. Then the total current I when its figure of merit is $K$ will beView Solution