SATHEE: Magnetics 4 Question 1

Magnetics 4 Question 1

1. Two magnetic dipoles $X$ and $Y$ are placed at a separation $d$ , with their axes perpendicular to each other. The dipole moment of $Y$ is twice that of $X$ . A

particle of charge $q$ is passing $(2 M)$ through their mid-point $P$ , at angle $θ = 45^{\circ}$ with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant? ( $d$ is much larger than the dimensions of the dipole)

(2019 Main, 8 April II)

(a) $\frac{μ_{0}}{4 π} \frac{M}{\frac{d^{3}}{2}} \times q v$

(b) 0

(d) $\frac{μ_{0}}{4 π} \frac{2 M}{\frac{d^{3}}{2}} \times q v$

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Answer:

Correct Answer: 1. (b)

Solution:

Let $2 l_{1}$ and $2 l_{2}$ be the length of dipole $X$ and $Y$ , respectively. For dipole $X$ , point $P$ lies on its axial line. So, magnetic field strength at $P$ due to $X$ is

$B_{X} = \frac{μ_{0}}{4 π} \cdot \frac{2 M r}{{(r^{2} - l_{1}^{2})}^{2}}, along O P$

Here,

$r = \frac{d}{2}$

Also,

$d ≫ l_{1}$

$\Rightarrow | B_{X} | = \frac{μ_{0}}{4 π} \cdot \frac{2 M (d / 2)}{(d / 2)^{4}} = \frac{μ_{0}}{4 π} \frac{2 M}{(d / 2)^{3}}$

Similarly, for dipole $Y$ , point $P$ lies on its equatorial line. So, magnetic field strength at $P$ due to $Y$ is

$B_{Y} = \frac{μ_{0}}{4 π} \cdot \frac{2 M}{{(r^{2} + l_{2}^{2})}^{3 / 2}}, ($ along a line perpendicular to $O^{'} P)$

Here,

$r = \frac{d}{2}$

Also,

$d ≫ l_{2}$

$\Rightarrow$

$| B_{Y} | = \frac{μ_{0}}{4 π} \frac{2 M}{(d / 2)^{3}}$

Thus, the resultant magnetic field due to $X$ and $Y$ at $P$ is

Since,

$B_{net} = B_{X} + B_{Y}$

$| B_{Y} | = | B_{X} |$

Thus, the resultant magnetic field $(B_{net}$ ) at $P$ will be at $45^{\circ}$ with the horizontal.

This means, direction of $B_{net}$ and velocity of the charged particle is same.

$∴$ Force on the charged particle moving with velocity $v$ in the presence of magnetic field which is

$B = q (v \times B) = q | v | B | \sin θ$

where, $θ$ is the angle between $B$ and $v$

According to the above analysis, we get

$\begin{aligned} ∴ & θ & = 0 \\ ∴ & F & = 0 \end{aligned}$

Thus, magnitude of force on the particle at that instant is zero. Key Idea Magnetic dipole moment of a current carrying loop is $m = I A (A - m^{2})$

where, $I =$ current in loop and $A =$ area of loop.

Let the given square loop has side $a$ , then its magnetic dipole moment will be

$m = I a^{2}$

When square is converted into a circular loop of radius $r$ ,

Then, wire length will be same in both areas,

$\Rightarrow 4 a = 2 π r \Rightarrow r = \frac{4 a}{2 π} = \frac{2 a}{π}$

Hence, area of circular loop formed is, $A^{'} = π r^{2}$

$= π \frac{2 a^{2}}{π} = \frac{4 a^{2}}{π}$

Magnitude of magnetic dipole moment of circular loop will be

$m^{'} = I A^{'} = I \frac{4 a^{2}}{π}$

Ratio of magnetic dipole moments of both shapes is,

$\frac{m^{'}}{m} = \frac{I \cdot \frac{4 a^{2}}{π}}{I a^{2}} = \frac{4}{π} \Rightarrow m^{'} = \frac{4 m}{π} (A - m)$

Magnetics 4 Question 1

1. Two magnetic dipoles $X$ and $Y$ are placed at a separation $d$ , with their axes perpendicular to each other. The dipole moment of $Y$ is twice that of $X$ . A

Answer:

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Magnetics 4 Question 1

1. Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. Two magnetic dipoles $X$ and $Y$ are placed at a separation $d$ , with their axes perpendicular to each other. The dipole moment of $Y$ is twice that of $X$ . A