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Moving Charges & Magnetism

153793 Two wires of same lengths are shaped into a square and a circle. If they carry same current, then the ratio of their magnetic moments is

1

2 : π

2

π : 2

3

π : 4

4

4 : π

Explanation:

C Let, side of square wire $= a$
Radius of circular wire $= r$
Length (perimeter) of square wire $(l_{1}) = 4 a$
$a = \frac{l_{1}}{4}$
Now,
Area of square $(A_{1}) = a^{2}$
$A_{1} = {(\frac{l_{1}}{4})}^{2}$
$A_{1} = \frac{l_{1}^{2}}{16}$
Circumference of circular wire $(l_{2}) = 2 π r$
$r = \frac{l_{2}}{2 π}$
Moving Charges \& Magnetism
Now, Area of circular wire $(A_{2}) = π r^{2}$
$A_{2} = π {(\frac{l_{2}}{2 π})}^{2}$
$A_{2} = \frac{l_{2}^{2}}{4 π}$
We know that,
Magnetic moment $(M) = IA$
$∴ M_{1} = I_{1} A_{1}$ and $M_{2} = I_{2} A_{2}$
Given, same current in both wire-
$∴ I_{1} = I_{2} = I$
And Length $= l_{1} = l_{2} = l$
$\frac{M_{1}}{M_{2}} = \frac{{IA}_{1}}{{IA}_{2}} = \frac{A_{1}}{A_{2}}$
$\frac{M_{1}}{M_{2}} = \frac{l_{1}^{2} / 16}{l_{2}^{2} / 4 π} = \frac{l^{2}}{16} \times \frac{4 π}{l^{2}} {∵ l_{1} = l_{2} = l}$
$\frac{M_{1}}{M_{2}} = \frac{π}{4}$
$M_{1} : M_{2} = π : 4$

Manipal UGET-2015

Moving Charges & Magnetism

153794 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $B$ is in the plane of the coil. If due to current $I$ in the triangle, a torque $τ$ acts on it. The side $l$ of the triangle is

1

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

2

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

3

\frac{2}{\sqrt{3}} {(\frac{τ}{Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

B Given,
Side of equilateral triangle $= l$
Area of equilateral triangle $= \frac{\sqrt{3}}{4} l^{2}$
$∵$ Torque $(τ) =$ BiNA $\sin θ$
$τ = Bi \times 1 \times \frac{\sqrt{3}}{4} l^{2} \sin 90^{\circ} [∵ θ = 90^{\circ}, N = 1]$
$τ = \frac{\sqrt{3}}{4} Bi l^{2}$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 \sqrt{\frac{τ}{\sqrt{3} Bi}}$
$l = 2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}$

Manipal UGET-2015

Moving Charges & Magnetism

153795 Two very long straight wires $P$ and $Q$ carry currents of $10 A$ and $20 A$ respectively and are at $20 cm$ apart. If a third wire, $R$ of length 15 cm having a current of $10 A$ is placed in middle between them, the direction of current in all the three wires is the same. How much force will act on $R$ ?

1

3.0 \times 10^{- 5} N

towards

Q

2

3.0 \times 10^{- 5} N

towards

P

3

3.0 \times 10^{- 7} N

towards

Q

4

3.0 \times 10^{- 7} N

towards

P

Explanation:

A Given,
Current on wire $P (I_{1}) = 10 A$
Current on wire $Q (I_{2}) = 20 A$
Current on wire $R (I_{3}) = 10 A$
Distance between wire $P$ and $Q = 20 m$
Length of wire $R, (l) = 15 cm = 0.15 m$

Force on $R$ due to wire $P$ -
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 I_{1} I_{3}}{r} \times l$
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 10}{0.1} \times 0.15$
Force on $R$ due to wire $Q$ -
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 I_{2} I_{3}}{r} \times l$
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 20}{0.1} \times 0.15$
Net force on wire $R$ ,
$F_{net} = F_{2} - F_{1}$
$F_{net} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10}{0.1} \times 0.15 (20 - 10)$
$F_{net} = 10^{- 7} \times \frac{2 \times 10 \times 0.15 \times 10}{0.1}$
$F_{net} = 10^{- 7} \times \frac{30}{0.1}$
$F_{net} = 10^{- 7} \times 300$
$F_{net} = 3 \times 10^{- 5} N towards Q$

CG PET -2016

Moving Charges & Magnetism

153796 A linear conductor of length $40 cm$ and carrying current $3 A$ is placed in a uniform magnetic field of intensity 500 Gauss. If the conductor makes an angle $30^{\circ}$ with the direction of magnetic field, the force acting on it as

1

3 \times 10^{4} N

2

3 \times 10^{2} N

3

3 \times 10^{- 2} N

4 None of these

Explanation:

C Given,
$I = 3 A, Length (L) = 40 cm = 0.4 m$
$B = 500 Gauss = 500 \times 10^{- 4} T, θ = 30^{\circ}$
$∵ Force (F) = ILB \sin θ$
$F = 3 \times 0.4 \times 500 \times 10^{- 4} \sin 30^{\circ}$
$F = 3 \times 200 \times 10^{- 4} \times \frac{1}{2}$
$F = 3 \times 10^{- 2} N$

CG PET- 2012

Moving Charges & Magnetism

153798 A coil has 25 turns and $10 cm$ radius. If it carries a $4 A$ current, then magnetic field at its centre is

1

6.28 \times 10^{- 5} T

2

6.28 \times 10^{- 3} T

3

6.28 \times 10^{- 4} T

4

6.28 \times 10^{- 2} T

Explanation:

C Given,
Turn of coil $(N) = 25$ turns, Radius $(r) = 10 cm = 0.1 m$ , Current $(I) = 4 A$
We know that,
$Magnetic field (B) = \frac{μ_{0} NI}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times 25 \times 4}{2 \times 0.1} [∵ μ_{0} = 4 π \times 10^{- 7}]$
$B = \frac{200 π \times 10^{- 7}}{0.1}$
$B = 2000 π \times 10^{- 7}$
$B = 2 π \times 10^{- 4}$
$B = 6.28 \times 10^{- 4} T$

CG PET- 2007

Moving Charges & Magnetism

153793 Two wires of same lengths are shaped into a square and a circle. If they carry same current, then the ratio of their magnetic moments is

1

2 : π

2

π : 2

3

π : 4

4

4 : π

Explanation:

C Let, side of square wire $= a$
Radius of circular wire $= r$
Length (perimeter) of square wire $(l_{1}) = 4 a$
$a = \frac{l_{1}}{4}$
Now,
Area of square $(A_{1}) = a^{2}$
$A_{1} = {(\frac{l_{1}}{4})}^{2}$
$A_{1} = \frac{l_{1}^{2}}{16}$
Circumference of circular wire $(l_{2}) = 2 π r$
$r = \frac{l_{2}}{2 π}$
Moving Charges \& Magnetism
Now, Area of circular wire $(A_{2}) = π r^{2}$
$A_{2} = π {(\frac{l_{2}}{2 π})}^{2}$
$A_{2} = \frac{l_{2}^{2}}{4 π}$
We know that,
Magnetic moment $(M) = IA$
$∴ M_{1} = I_{1} A_{1}$ and $M_{2} = I_{2} A_{2}$
Given, same current in both wire-
$∴ I_{1} = I_{2} = I$
And Length $= l_{1} = l_{2} = l$
$\frac{M_{1}}{M_{2}} = \frac{{IA}_{1}}{{IA}_{2}} = \frac{A_{1}}{A_{2}}$
$\frac{M_{1}}{M_{2}} = \frac{l_{1}^{2} / 16}{l_{2}^{2} / 4 π} = \frac{l^{2}}{16} \times \frac{4 π}{l^{2}} {∵ l_{1} = l_{2} = l}$
$\frac{M_{1}}{M_{2}} = \frac{π}{4}$
$M_{1} : M_{2} = π : 4$

Manipal UGET-2015

Moving Charges & Magnetism

153794 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $B$ is in the plane of the coil. If due to current $I$ in the triangle, a torque $τ$ acts on it. The side $l$ of the triangle is

1

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

2

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

3

\frac{2}{\sqrt{3}} {(\frac{τ}{Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

B Given,
Side of equilateral triangle $= l$
Area of equilateral triangle $= \frac{\sqrt{3}}{4} l^{2}$
$∵$ Torque $(τ) =$ BiNA $\sin θ$
$τ = Bi \times 1 \times \frac{\sqrt{3}}{4} l^{2} \sin 90^{\circ} [∵ θ = 90^{\circ}, N = 1]$
$τ = \frac{\sqrt{3}}{4} Bi l^{2}$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 \sqrt{\frac{τ}{\sqrt{3} Bi}}$
$l = 2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}$

Manipal UGET-2015

Moving Charges & Magnetism

153795 Two very long straight wires $P$ and $Q$ carry currents of $10 A$ and $20 A$ respectively and are at $20 cm$ apart. If a third wire, $R$ of length 15 cm having a current of $10 A$ is placed in middle between them, the direction of current in all the three wires is the same. How much force will act on $R$ ?

1

3.0 \times 10^{- 5} N

towards

Q

2

3.0 \times 10^{- 5} N

towards

P

3

3.0 \times 10^{- 7} N

towards

Q

4

3.0 \times 10^{- 7} N

towards

P

Explanation:

A Given,
Current on wire $P (I_{1}) = 10 A$
Current on wire $Q (I_{2}) = 20 A$
Current on wire $R (I_{3}) = 10 A$
Distance between wire $P$ and $Q = 20 m$
Length of wire $R, (l) = 15 cm = 0.15 m$

Force on $R$ due to wire $P$ -
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 I_{1} I_{3}}{r} \times l$
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 10}{0.1} \times 0.15$
Force on $R$ due to wire $Q$ -
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 I_{2} I_{3}}{r} \times l$
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 20}{0.1} \times 0.15$
Net force on wire $R$ ,
$F_{net} = F_{2} - F_{1}$
$F_{net} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10}{0.1} \times 0.15 (20 - 10)$
$F_{net} = 10^{- 7} \times \frac{2 \times 10 \times 0.15 \times 10}{0.1}$
$F_{net} = 10^{- 7} \times \frac{30}{0.1}$
$F_{net} = 10^{- 7} \times 300$
$F_{net} = 3 \times 10^{- 5} N towards Q$

CG PET -2016

Moving Charges & Magnetism

153796 A linear conductor of length $40 cm$ and carrying current $3 A$ is placed in a uniform magnetic field of intensity 500 Gauss. If the conductor makes an angle $30^{\circ}$ with the direction of magnetic field, the force acting on it as

1

3 \times 10^{4} N

2

3 \times 10^{2} N

3

3 \times 10^{- 2} N

4 None of these

Explanation:

C Given,
$I = 3 A, Length (L) = 40 cm = 0.4 m$
$B = 500 Gauss = 500 \times 10^{- 4} T, θ = 30^{\circ}$
$∵ Force (F) = ILB \sin θ$
$F = 3 \times 0.4 \times 500 \times 10^{- 4} \sin 30^{\circ}$
$F = 3 \times 200 \times 10^{- 4} \times \frac{1}{2}$
$F = 3 \times 10^{- 2} N$

CG PET- 2012

Moving Charges & Magnetism

153798 A coil has 25 turns and $10 cm$ radius. If it carries a $4 A$ current, then magnetic field at its centre is

1

6.28 \times 10^{- 5} T

2

6.28 \times 10^{- 3} T

3

6.28 \times 10^{- 4} T

4

6.28 \times 10^{- 2} T

Explanation:

C Given,
Turn of coil $(N) = 25$ turns, Radius $(r) = 10 cm = 0.1 m$ , Current $(I) = 4 A$
We know that,
$Magnetic field (B) = \frac{μ_{0} NI}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times 25 \times 4}{2 \times 0.1} [∵ μ_{0} = 4 π \times 10^{- 7}]$
$B = \frac{200 π \times 10^{- 7}}{0.1}$
$B = 2000 π \times 10^{- 7}$
$B = 2 π \times 10^{- 4}$
$B = 6.28 \times 10^{- 4} T$

CG PET- 2007

Moving Charges & Magnetism

153793 Two wires of same lengths are shaped into a square and a circle. If they carry same current, then the ratio of their magnetic moments is

1

2 : π

2

π : 2

3

π : 4

4

4 : π

Explanation:

C Let, side of square wire $= a$
Radius of circular wire $= r$
Length (perimeter) of square wire $(l_{1}) = 4 a$
$a = \frac{l_{1}}{4}$
Now,
Area of square $(A_{1}) = a^{2}$
$A_{1} = {(\frac{l_{1}}{4})}^{2}$
$A_{1} = \frac{l_{1}^{2}}{16}$
Circumference of circular wire $(l_{2}) = 2 π r$
$r = \frac{l_{2}}{2 π}$
Moving Charges \& Magnetism
Now, Area of circular wire $(A_{2}) = π r^{2}$
$A_{2} = π {(\frac{l_{2}}{2 π})}^{2}$
$A_{2} = \frac{l_{2}^{2}}{4 π}$
We know that,
Magnetic moment $(M) = IA$
$∴ M_{1} = I_{1} A_{1}$ and $M_{2} = I_{2} A_{2}$
Given, same current in both wire-
$∴ I_{1} = I_{2} = I$
And Length $= l_{1} = l_{2} = l$
$\frac{M_{1}}{M_{2}} = \frac{{IA}_{1}}{{IA}_{2}} = \frac{A_{1}}{A_{2}}$
$\frac{M_{1}}{M_{2}} = \frac{l_{1}^{2} / 16}{l_{2}^{2} / 4 π} = \frac{l^{2}}{16} \times \frac{4 π}{l^{2}} {∵ l_{1} = l_{2} = l}$
$\frac{M_{1}}{M_{2}} = \frac{π}{4}$
$M_{1} : M_{2} = π : 4$

Manipal UGET-2015

Moving Charges & Magnetism

153794 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $B$ is in the plane of the coil. If due to current $I$ in the triangle, a torque $τ$ acts on it. The side $l$ of the triangle is

1

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

2

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

3

\frac{2}{\sqrt{3}} {(\frac{τ}{Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

B Given,
Side of equilateral triangle $= l$
Area of equilateral triangle $= \frac{\sqrt{3}}{4} l^{2}$
$∵$ Torque $(τ) =$ BiNA $\sin θ$
$τ = Bi \times 1 \times \frac{\sqrt{3}}{4} l^{2} \sin 90^{\circ} [∵ θ = 90^{\circ}, N = 1]$
$τ = \frac{\sqrt{3}}{4} Bi l^{2}$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 \sqrt{\frac{τ}{\sqrt{3} Bi}}$
$l = 2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}$

Manipal UGET-2015

Moving Charges & Magnetism

153795 Two very long straight wires $P$ and $Q$ carry currents of $10 A$ and $20 A$ respectively and are at $20 cm$ apart. If a third wire, $R$ of length 15 cm having a current of $10 A$ is placed in middle between them, the direction of current in all the three wires is the same. How much force will act on $R$ ?

1

3.0 \times 10^{- 5} N

towards

Q

2

3.0 \times 10^{- 5} N

towards

P

3

3.0 \times 10^{- 7} N

towards

Q

4

3.0 \times 10^{- 7} N

towards

P

Explanation:

A Given,
Current on wire $P (I_{1}) = 10 A$
Current on wire $Q (I_{2}) = 20 A$
Current on wire $R (I_{3}) = 10 A$
Distance between wire $P$ and $Q = 20 m$
Length of wire $R, (l) = 15 cm = 0.15 m$

Force on $R$ due to wire $P$ -
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 I_{1} I_{3}}{r} \times l$
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 10}{0.1} \times 0.15$
Force on $R$ due to wire $Q$ -
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 I_{2} I_{3}}{r} \times l$
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 20}{0.1} \times 0.15$
Net force on wire $R$ ,
$F_{net} = F_{2} - F_{1}$
$F_{net} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10}{0.1} \times 0.15 (20 - 10)$
$F_{net} = 10^{- 7} \times \frac{2 \times 10 \times 0.15 \times 10}{0.1}$
$F_{net} = 10^{- 7} \times \frac{30}{0.1}$
$F_{net} = 10^{- 7} \times 300$
$F_{net} = 3 \times 10^{- 5} N towards Q$

CG PET -2016

Moving Charges & Magnetism

153796 A linear conductor of length $40 cm$ and carrying current $3 A$ is placed in a uniform magnetic field of intensity 500 Gauss. If the conductor makes an angle $30^{\circ}$ with the direction of magnetic field, the force acting on it as

1

3 \times 10^{4} N

2

3 \times 10^{2} N

3

3 \times 10^{- 2} N

4 None of these

Explanation:

C Given,
$I = 3 A, Length (L) = 40 cm = 0.4 m$
$B = 500 Gauss = 500 \times 10^{- 4} T, θ = 30^{\circ}$
$∵ Force (F) = ILB \sin θ$
$F = 3 \times 0.4 \times 500 \times 10^{- 4} \sin 30^{\circ}$
$F = 3 \times 200 \times 10^{- 4} \times \frac{1}{2}$
$F = 3 \times 10^{- 2} N$

CG PET- 2012

Moving Charges & Magnetism

153798 A coil has 25 turns and $10 cm$ radius. If it carries a $4 A$ current, then magnetic field at its centre is

1

6.28 \times 10^{- 5} T

2

6.28 \times 10^{- 3} T

3

6.28 \times 10^{- 4} T

4

6.28 \times 10^{- 2} T

Explanation:

C Given,
Turn of coil $(N) = 25$ turns, Radius $(r) = 10 cm = 0.1 m$ , Current $(I) = 4 A$
We know that,
$Magnetic field (B) = \frac{μ_{0} NI}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times 25 \times 4}{2 \times 0.1} [∵ μ_{0} = 4 π \times 10^{- 7}]$
$B = \frac{200 π \times 10^{- 7}}{0.1}$
$B = 2000 π \times 10^{- 7}$
$B = 2 π \times 10^{- 4}$
$B = 6.28 \times 10^{- 4} T$

CG PET- 2007

Moving Charges & Magnetism

153793 Two wires of same lengths are shaped into a square and a circle. If they carry same current, then the ratio of their magnetic moments is

1

2 : π

2

π : 2

3

π : 4

4

4 : π

Explanation:

C Let, side of square wire $= a$
Radius of circular wire $= r$
Length (perimeter) of square wire $(l_{1}) = 4 a$
$a = \frac{l_{1}}{4}$
Now,
Area of square $(A_{1}) = a^{2}$
$A_{1} = {(\frac{l_{1}}{4})}^{2}$
$A_{1} = \frac{l_{1}^{2}}{16}$
Circumference of circular wire $(l_{2}) = 2 π r$
$r = \frac{l_{2}}{2 π}$
Moving Charges \& Magnetism
Now, Area of circular wire $(A_{2}) = π r^{2}$
$A_{2} = π {(\frac{l_{2}}{2 π})}^{2}$
$A_{2} = \frac{l_{2}^{2}}{4 π}$
We know that,
Magnetic moment $(M) = IA$
$∴ M_{1} = I_{1} A_{1}$ and $M_{2} = I_{2} A_{2}$
Given, same current in both wire-
$∴ I_{1} = I_{2} = I$
And Length $= l_{1} = l_{2} = l$
$\frac{M_{1}}{M_{2}} = \frac{{IA}_{1}}{{IA}_{2}} = \frac{A_{1}}{A_{2}}$
$\frac{M_{1}}{M_{2}} = \frac{l_{1}^{2} / 16}{l_{2}^{2} / 4 π} = \frac{l^{2}}{16} \times \frac{4 π}{l^{2}} {∵ l_{1} = l_{2} = l}$
$\frac{M_{1}}{M_{2}} = \frac{π}{4}$
$M_{1} : M_{2} = π : 4$

Manipal UGET-2015

Moving Charges & Magnetism

153794 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $B$ is in the plane of the coil. If due to current $I$ in the triangle, a torque $τ$ acts on it. The side $l$ of the triangle is

1

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

2

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

3

\frac{2}{\sqrt{3}} {(\frac{τ}{Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

B Given,
Side of equilateral triangle $= l$
Area of equilateral triangle $= \frac{\sqrt{3}}{4} l^{2}$
$∵$ Torque $(τ) =$ BiNA $\sin θ$
$τ = Bi \times 1 \times \frac{\sqrt{3}}{4} l^{2} \sin 90^{\circ} [∵ θ = 90^{\circ}, N = 1]$
$τ = \frac{\sqrt{3}}{4} Bi l^{2}$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 \sqrt{\frac{τ}{\sqrt{3} Bi}}$
$l = 2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}$

Manipal UGET-2015

Moving Charges & Magnetism

153795 Two very long straight wires $P$ and $Q$ carry currents of $10 A$ and $20 A$ respectively and are at $20 cm$ apart. If a third wire, $R$ of length 15 cm having a current of $10 A$ is placed in middle between them, the direction of current in all the three wires is the same. How much force will act on $R$ ?

1

3.0 \times 10^{- 5} N

towards

Q

2

3.0 \times 10^{- 5} N

towards

P

3

3.0 \times 10^{- 7} N

towards

Q

4

3.0 \times 10^{- 7} N

towards

P

Explanation:

A Given,
Current on wire $P (I_{1}) = 10 A$
Current on wire $Q (I_{2}) = 20 A$
Current on wire $R (I_{3}) = 10 A$
Distance between wire $P$ and $Q = 20 m$
Length of wire $R, (l) = 15 cm = 0.15 m$

Force on $R$ due to wire $P$ -
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 I_{1} I_{3}}{r} \times l$
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 10}{0.1} \times 0.15$
Force on $R$ due to wire $Q$ -
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 I_{2} I_{3}}{r} \times l$
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 20}{0.1} \times 0.15$
Net force on wire $R$ ,
$F_{net} = F_{2} - F_{1}$
$F_{net} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10}{0.1} \times 0.15 (20 - 10)$
$F_{net} = 10^{- 7} \times \frac{2 \times 10 \times 0.15 \times 10}{0.1}$
$F_{net} = 10^{- 7} \times \frac{30}{0.1}$
$F_{net} = 10^{- 7} \times 300$
$F_{net} = 3 \times 10^{- 5} N towards Q$

CG PET -2016

Moving Charges & Magnetism

153796 A linear conductor of length $40 cm$ and carrying current $3 A$ is placed in a uniform magnetic field of intensity 500 Gauss. If the conductor makes an angle $30^{\circ}$ with the direction of magnetic field, the force acting on it as

1

3 \times 10^{4} N

2

3 \times 10^{2} N

3

3 \times 10^{- 2} N

4 None of these

Explanation:

C Given,
$I = 3 A, Length (L) = 40 cm = 0.4 m$
$B = 500 Gauss = 500 \times 10^{- 4} T, θ = 30^{\circ}$
$∵ Force (F) = ILB \sin θ$
$F = 3 \times 0.4 \times 500 \times 10^{- 4} \sin 30^{\circ}$
$F = 3 \times 200 \times 10^{- 4} \times \frac{1}{2}$
$F = 3 \times 10^{- 2} N$

CG PET- 2012

Moving Charges & Magnetism

153798 A coil has 25 turns and $10 cm$ radius. If it carries a $4 A$ current, then magnetic field at its centre is

1

6.28 \times 10^{- 5} T

2

6.28 \times 10^{- 3} T

3

6.28 \times 10^{- 4} T

4

6.28 \times 10^{- 2} T

Explanation:

C Given,
Turn of coil $(N) = 25$ turns, Radius $(r) = 10 cm = 0.1 m$ , Current $(I) = 4 A$
We know that,
$Magnetic field (B) = \frac{μ_{0} NI}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times 25 \times 4}{2 \times 0.1} [∵ μ_{0} = 4 π \times 10^{- 7}]$
$B = \frac{200 π \times 10^{- 7}}{0.1}$
$B = 2000 π \times 10^{- 7}$
$B = 2 π \times 10^{- 4}$
$B = 6.28 \times 10^{- 4} T$

CG PET- 2007

Moving Charges & Magnetism

153793 Two wires of same lengths are shaped into a square and a circle. If they carry same current, then the ratio of their magnetic moments is

1

2 : π

2

π : 2

3

π : 4

4

4 : π

Explanation:

C Let, side of square wire $= a$
Radius of circular wire $= r$
Length (perimeter) of square wire $(l_{1}) = 4 a$
$a = \frac{l_{1}}{4}$
Now,
Area of square $(A_{1}) = a^{2}$
$A_{1} = {(\frac{l_{1}}{4})}^{2}$
$A_{1} = \frac{l_{1}^{2}}{16}$
Circumference of circular wire $(l_{2}) = 2 π r$
$r = \frac{l_{2}}{2 π}$
Moving Charges \& Magnetism
Now, Area of circular wire $(A_{2}) = π r^{2}$
$A_{2} = π {(\frac{l_{2}}{2 π})}^{2}$
$A_{2} = \frac{l_{2}^{2}}{4 π}$
We know that,
Magnetic moment $(M) = IA$
$∴ M_{1} = I_{1} A_{1}$ and $M_{2} = I_{2} A_{2}$
Given, same current in both wire-
$∴ I_{1} = I_{2} = I$
And Length $= l_{1} = l_{2} = l$
$\frac{M_{1}}{M_{2}} = \frac{{IA}_{1}}{{IA}_{2}} = \frac{A_{1}}{A_{2}}$
$\frac{M_{1}}{M_{2}} = \frac{l_{1}^{2} / 16}{l_{2}^{2} / 4 π} = \frac{l^{2}}{16} \times \frac{4 π}{l^{2}} {∵ l_{1} = l_{2} = l}$
$\frac{M_{1}}{M_{2}} = \frac{π}{4}$
$M_{1} : M_{2} = π : 4$

Manipal UGET-2015

Moving Charges & Magnetism

153794 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $B$ is in the plane of the coil. If due to current $I$ in the triangle, a torque $τ$ acts on it. The side $l$ of the triangle is

1

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

2

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

3

\frac{2}{\sqrt{3}} {(\frac{τ}{Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

B Given,
Side of equilateral triangle $= l$
Area of equilateral triangle $= \frac{\sqrt{3}}{4} l^{2}$
$∵$ Torque $(τ) =$ BiNA $\sin θ$
$τ = Bi \times 1 \times \frac{\sqrt{3}}{4} l^{2} \sin 90^{\circ} [∵ θ = 90^{\circ}, N = 1]$
$τ = \frac{\sqrt{3}}{4} Bi l^{2}$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 \sqrt{\frac{τ}{\sqrt{3} Bi}}$
$l = 2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}$

Manipal UGET-2015

Moving Charges & Magnetism

153795 Two very long straight wires $P$ and $Q$ carry currents of $10 A$ and $20 A$ respectively and are at $20 cm$ apart. If a third wire, $R$ of length 15 cm having a current of $10 A$ is placed in middle between them, the direction of current in all the three wires is the same. How much force will act on $R$ ?

1

3.0 \times 10^{- 5} N

towards

Q

2

3.0 \times 10^{- 5} N

towards

P

3

3.0 \times 10^{- 7} N

towards

Q

4

3.0 \times 10^{- 7} N

towards

P

Explanation:

A Given,
Current on wire $P (I_{1}) = 10 A$
Current on wire $Q (I_{2}) = 20 A$
Current on wire $R (I_{3}) = 10 A$
Distance between wire $P$ and $Q = 20 m$
Length of wire $R, (l) = 15 cm = 0.15 m$

Force on $R$ due to wire $P$ -
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 I_{1} I_{3}}{r} \times l$
$F_{1} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 10}{0.1} \times 0.15$
Force on $R$ due to wire $Q$ -
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 I_{2} I_{3}}{r} \times l$
$F_{2} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10 \times 20}{0.1} \times 0.15$
Net force on wire $R$ ,
$F_{net} = F_{2} - F_{1}$
$F_{net} = \frac{μ_{0}}{4 π} \times \frac{2 \times 10}{0.1} \times 0.15 (20 - 10)$
$F_{net} = 10^{- 7} \times \frac{2 \times 10 \times 0.15 \times 10}{0.1}$
$F_{net} = 10^{- 7} \times \frac{30}{0.1}$
$F_{net} = 10^{- 7} \times 300$
$F_{net} = 3 \times 10^{- 5} N towards Q$

CG PET -2016

Moving Charges & Magnetism

153796 A linear conductor of length $40 cm$ and carrying current $3 A$ is placed in a uniform magnetic field of intensity 500 Gauss. If the conductor makes an angle $30^{\circ}$ with the direction of magnetic field, the force acting on it as

1

3 \times 10^{4} N

2

3 \times 10^{2} N

3

3 \times 10^{- 2} N

4 None of these

Explanation:

C Given,
$I = 3 A, Length (L) = 40 cm = 0.4 m$
$B = 500 Gauss = 500 \times 10^{- 4} T, θ = 30^{\circ}$
$∵ Force (F) = ILB \sin θ$
$F = 3 \times 0.4 \times 500 \times 10^{- 4} \sin 30^{\circ}$
$F = 3 \times 200 \times 10^{- 4} \times \frac{1}{2}$
$F = 3 \times 10^{- 2} N$

CG PET- 2012

Moving Charges & Magnetism

153798 A coil has 25 turns and $10 cm$ radius. If it carries a $4 A$ current, then magnetic field at its centre is

1

6.28 \times 10^{- 5} T

2

6.28 \times 10^{- 3} T

3

6.28 \times 10^{- 4} T

4

6.28 \times 10^{- 2} T

Explanation:

C Given,
Turn of coil $(N) = 25$ turns, Radius $(r) = 10 cm = 0.1 m$ , Current $(I) = 4 A$
We know that,
$Magnetic field (B) = \frac{μ_{0} NI}{2 r}$
$B = \frac{4 π \times 10^{- 7} \times 25 \times 4}{2 \times 0.1} [∵ μ_{0} = 4 π \times 10^{- 7}]$
$B = \frac{200 π \times 10^{- 7}}{0.1}$
$B = 2000 π \times 10^{- 7}$
$B = 2 π \times 10^{- 4}$
$B = 6.28 \times 10^{- 4} T$

CG PET- 2007