A coil of 12 turns made by a constant length | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\mathrm{B} \propto \mathrm{N}^{2} \Rightarrow \frac{\mathrm{B}_{2}}{\mathrm{B}_{1}}=\left(\frac{\mathrm{N}_{2}}{\mathrm{N}_{1}}\right)^{2}=\left(\fra

Vedclass · Accepted Answer

$93.75\%$ decreases

Vedclass · Answer

$\mathrm{B} \propto \mathrm{N}^{2} \Rightarrow \frac{\mathrm{B}_{2}}{\mathrm{B}_{1}}=\left(\frac{\mathrm{N}_{2}}{\mathrm{N}_{1}}ight)^{2}=\left(\frac{3}{12}ight)^{2}=\frac{9}{144}$

$\Rightarrow \mathrm{B}_{2}=\frac{9}{144} \mathrm{B}_{1}$

$\%=\frac{\Delta \mathrm{B}}{\mathrm{B}_{1}} 	imes 100=\frac{\mathrm{B}_{1}-\mathrm{B}_{2}}{\mathrm{B}_{1}} 	imes 100=\frac{135}{144} 	imes 100$

$\quad=93.75 \%$

A coil of $12$ turns made by a constant length current carrying wire. If number of turns makes $3$ then change in magnetic field produced at its centre

Similar Questions

A coil of $50\, turns$ and $4\,cm$ radius carries $2\,A$ current then magnetic field at its centre is......$mT$

In the figure, what is the magnetic field at the point $O$

If an electron revolves around a nucleus in a circular orbit of radius $R$ with frequency $n$, then the magnetic field produced at the centre of the nucleus will be

When a certain length of wire is turned into one circular loop, the magnetic induction at the centre of coil due to some current flowing is ${B_1}$ If the same wire is turned into three loops to make a circular coil, the magnetic induction at the center of this coil for the same current will be

The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is ${B_1}$. The field at the centre of another coil of radius $2 r$ carrying same current $I$ is ${B_2}$. The ratio $\frac{{{B_1}}}{{{B_2}}}$ is