(B) The magnetic field at the center of a circular coil is given by the formula $B = \frac{\mu_0 NI}{2R}$.Since the current $I$ and the radius $R$ are

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(B) The magnetic field at the center of a circular coil is given by the formula $B = \frac{\mu_0 NI}{2R}$.Since the current $I$ and the radius $R$ are the same for both coils,the magnetic field $B$ is directly proportional to the number of turns $N$,i.e.,$B \propto N$.Therefore,the ratio of the magnetic fields is $\frac{B_{x}}{B_{y}} = \frac{N_{x}}{N_{y}}$.Given $N_{x} = 200$ and $N_{y} = 400$,we have $\frac{B_{x}}{B_{y}} = \frac{200}{400} = \frac{1}{2}$.Thus,the ratio is $1 : 2$.

Class 12 Physics Moving Charges and Magnetism Biot-Savart's Law and its application

Medium

$B_{x}$ and $B_{y}$ are the magnetic fields at the center of two coils $x$ and $y$ respectively,each carrying equal current. If coil $x$ has $200$ turns and $20 \ cm$ radius,and coil $y$ has $400$ turns and $20 \ cm$ radius,the ratio of $B_{x}$ and $B_{y}$ is $:-$

A
$1 : 1$
B
$1 : 2$
C
$2 : 1$
D
$4 : 1$

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Class 12 Questions

Class 12

Physics Questions

Chapter

Moving Charges and Magnetism

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Biot-Savart's Law and its application

The magnetic induction at the centre of a current-carrying circular coil of radius $r$ is

Medium

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$A$ charged particle $A$ with charge $q = 2 \, C$ has a velocity $v = 100 \, m/s$. When it passes through point $A$ with the velocity in the direction shown in the figure,the strength of the magnetic field at point $B$ due to this moving charge is ....... $\mu T$ $(r = 2 \, m)$.

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$A$ horizontal overhead power line is at a height of $4\,m$ from the ground and carries a current of $100\,A$ from east to west. The magnetic field directly below it on the ground is $(\mu _0 = 4\pi \times 10^{-7}\,TmA^{-1})$

Medium

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$A$ cylindrical conductor of radius $R$ carries a current $i$. The value of the magnetic field at a point which is $R/4$ distance inside from the surface is $10 \, T$. Find the value of the magnetic field at a point which is $4R$ distance outside from the surface.

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Write the formula for the magnetic field due to a circular current-carrying loop having $N$ turns and $R$ radius at a point on the axis of the loop at a distance $x$ from the center.

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$B_{x}$ and $B_{y}$ are the magnetic fields at the center of two coils $x$ and $y$ respectively,each carrying equal current. If coil $x$ has $200$ turns and $20 \ cm$ radius,and coil $y$ has $400$ turns and $20 \ cm$ radius,the ratio of $B_{x}$ and $B_{y}$ is $:-$

Similar Questions

The magnetic induction at the centre of a current-carrying circular coil of radius $r$ is

$A$ charged particle $A$ with charge $q = 2 \, C$ has a velocity $v = 100 \, m/s$. When it passes through point $A$ with the velocity in the direction shown in the figure,the strength of the magnetic field at point $B$ due to this moving charge is ....... $\mu T$ $(r = 2 \, m)$.

$A$ horizontal overhead power line is at a height of $4\,m$ from the ground and carries a current of $100\,A$ from east to west. The magnetic field directly below it on the ground is $(\mu _0 = 4\pi \times 10^{-7}\,TmA^{-1})$

$A$ cylindrical conductor of radius $R$ carries a current $i$. The value of the magnetic field at a point which is $R/4$ distance inside from the surface is $10 \, T$. Find the value of the magnetic field at a point which is $4R$ distance outside from the surface.

Write the formula for the magnetic field due to a circular current-carrying loop having $N$ turns and $R$ radius at a point on the axis of the loop at a distance $x$ from the center.

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