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Moving Charges & Magnetism

153844 A current carrying closed loop in the form of a right angled isosceles $△ ABC$ is placed in a uniform magnetic field acting along $AB$ . If the magnetic force on the $arm B C$ is $F$ , the force on the arm $AC$ is

1

- F

2

F

3

\sqrt{2 F}

4

- \sqrt{2 F}

Explanation:

A Force along $AB$ is zero as magnetic field is along $AB$ . The net force on a current carrying loop is always zero.
$F_{net} = F_{AB} + F_{BC} + F_{AC}$
$F_{AB} = 0 and F_{BC} = F$
Then, $0 = 0 + F + F_{AC}$
$F_{AC} = - F$

CBESE AIPMT 2011

Moving Charges & Magnetism

153845 A square current carrying loop is suspended in a uniform magnetic field acting in the place of the loop. If the force on one arm of the loop is $F$ , the net force on the remaining three arms of the loop is

1

3 F

2

- F

3

- 3 F

4

F

Explanation:

B :Consider a square PQRS which is suspended in a uniform magnetic field $B$ and force on one arm of the loop is $F$ .

As shown from the figure, side QR and PS are parallel to the direction of magnetic field. The force on these arm two arms of parallel to the field is zero.
$\vec{F} = i f (\vec{d} l \times \vec{B})$
$\vec{d} l =$ displacement and $\vec{B} =$ magnetic field
Hence, $F_{QR} = F_{PS} = 0$
Net force acting on the close loop
$F_{PQ} + F_{QR} + F_{RS} + F_{PS} = 0$
$F_{PQ} + 0 + F_{RS} + 0 = 0$
$F_{PQ} = - F_{RS}$
$F = - F_{RS}$
$F_{RS} = - F$

CBESE AIPMT 2010

Moving Charges & Magnetism

153846 A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are $F_{1}, F_{2}$ and $F_{3}$ respectively and are in the plane of the paper and along the directions shown in figure, the force on the segment $Q P$ is

1

F_{3} - F_{1} - F_{2}

2

\sqrt{{(F_{3} - F_{1})}^{2} + F_{2}^{2}}

3

\sqrt{{(F_{3} - F_{1})}^{2} - F_{2}^{2}}

4

F_{3} - F_{1} + F_{2}

Explanation:

B Since, it is placed in a uniform magnetic field.
Therefore, the total force on the current carrying in closed loop will be 0 .

Net horizontal force -
$F_{H} = F_{3} - F_{1}$
Net vertical force -
$F_{v} = F_{2}$
$∴$ Force on the segment QP is-
$| \vec{F} | = \sqrt{{(F_{3} - F_{1})}^{2} + F_{2}^{2}}$

CBESE AIPMT 2008

Moving Charges & Magnetism

153847 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $F$ is in plane of the coil. If due to a 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})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

3

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

C The current flowing clockwise in an equilateral triangle has a magnetic field in the direction of $\hat{k}$ .
$τ = BiNA \sin θ$
$τ = BiNA \sin 90^{\circ}$
$∴ τ = Bi \times \frac{\sqrt{3}}{4} l^{2} \times 1$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 {(\frac{τ}{Bi \sqrt{3}})}^{\frac{1}{2}}$

CBESE AIPMT 2005

Moving Charges & Magnetism

153844 A current carrying closed loop in the form of a right angled isosceles $△ ABC$ is placed in a uniform magnetic field acting along $AB$ . If the magnetic force on the $arm B C$ is $F$ , the force on the arm $AC$ is

1

- F

2

F

3

\sqrt{2 F}

4

- \sqrt{2 F}

Explanation:

A Force along $AB$ is zero as magnetic field is along $AB$ . The net force on a current carrying loop is always zero.
$F_{net} = F_{AB} + F_{BC} + F_{AC}$
$F_{AB} = 0 and F_{BC} = F$
Then, $0 = 0 + F + F_{AC}$
$F_{AC} = - F$

CBESE AIPMT 2011

Moving Charges & Magnetism

153845 A square current carrying loop is suspended in a uniform magnetic field acting in the place of the loop. If the force on one arm of the loop is $F$ , the net force on the remaining three arms of the loop is

1

3 F

2

- F

3

- 3 F

4

F

Explanation:

B :Consider a square PQRS which is suspended in a uniform magnetic field $B$ and force on one arm of the loop is $F$ .

As shown from the figure, side QR and PS are parallel to the direction of magnetic field. The force on these arm two arms of parallel to the field is zero.
$\vec{F} = i f (\vec{d} l \times \vec{B})$
$\vec{d} l =$ displacement and $\vec{B} =$ magnetic field
Hence, $F_{QR} = F_{PS} = 0$
Net force acting on the close loop
$F_{PQ} + F_{QR} + F_{RS} + F_{PS} = 0$
$F_{PQ} + 0 + F_{RS} + 0 = 0$
$F_{PQ} = - F_{RS}$
$F = - F_{RS}$
$F_{RS} = - F$

CBESE AIPMT 2010

Moving Charges & Magnetism

153846 A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are $F_{1}, F_{2}$ and $F_{3}$ respectively and are in the plane of the paper and along the directions shown in figure, the force on the segment $Q P$ is

1

F_{3} - F_{1} - F_{2}

2

\sqrt{{(F_{3} - F_{1})}^{2} + F_{2}^{2}}

3

\sqrt{{(F_{3} - F_{1})}^{2} - F_{2}^{2}}

4

F_{3} - F_{1} + F_{2}

Explanation:

B Since, it is placed in a uniform magnetic field.
Therefore, the total force on the current carrying in closed loop will be 0 .

Net horizontal force -
$F_{H} = F_{3} - F_{1}$
Net vertical force -
$F_{v} = F_{2}$
$∴$ Force on the segment QP is-
$| \vec{F} | = \sqrt{{(F_{3} - F_{1})}^{2} + F_{2}^{2}}$

CBESE AIPMT 2008

Moving Charges & Magnetism

153847 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $F$ is in plane of the coil. If due to a 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})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

3

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

C The current flowing clockwise in an equilateral triangle has a magnetic field in the direction of $\hat{k}$ .
$τ = BiNA \sin θ$
$τ = BiNA \sin 90^{\circ}$
$∴ τ = Bi \times \frac{\sqrt{3}}{4} l^{2} \times 1$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 {(\frac{τ}{Bi \sqrt{3}})}^{\frac{1}{2}}$

CBESE AIPMT 2005

Moving Charges & Magnetism

153844 A current carrying closed loop in the form of a right angled isosceles $△ ABC$ is placed in a uniform magnetic field acting along $AB$ . If the magnetic force on the $arm B C$ is $F$ , the force on the arm $AC$ is

1

- F

2

F

3

\sqrt{2 F}

4

- \sqrt{2 F}

Explanation:

A Force along $AB$ is zero as magnetic field is along $AB$ . The net force on a current carrying loop is always zero.
$F_{net} = F_{AB} + F_{BC} + F_{AC}$
$F_{AB} = 0 and F_{BC} = F$
Then, $0 = 0 + F + F_{AC}$
$F_{AC} = - F$

CBESE AIPMT 2011

Moving Charges & Magnetism

153845 A square current carrying loop is suspended in a uniform magnetic field acting in the place of the loop. If the force on one arm of the loop is $F$ , the net force on the remaining three arms of the loop is

1

3 F

2

- F

3

- 3 F

4

F

Explanation:

B :Consider a square PQRS which is suspended in a uniform magnetic field $B$ and force on one arm of the loop is $F$ .

As shown from the figure, side QR and PS are parallel to the direction of magnetic field. The force on these arm two arms of parallel to the field is zero.
$\vec{F} = i f (\vec{d} l \times \vec{B})$
$\vec{d} l =$ displacement and $\vec{B} =$ magnetic field
Hence, $F_{QR} = F_{PS} = 0$
Net force acting on the close loop
$F_{PQ} + F_{QR} + F_{RS} + F_{PS} = 0$
$F_{PQ} + 0 + F_{RS} + 0 = 0$
$F_{PQ} = - F_{RS}$
$F = - F_{RS}$
$F_{RS} = - F$

CBESE AIPMT 2010

Moving Charges & Magnetism

153846 A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are $F_{1}, F_{2}$ and $F_{3}$ respectively and are in the plane of the paper and along the directions shown in figure, the force on the segment $Q P$ is

1

F_{3} - F_{1} - F_{2}

2

\sqrt{{(F_{3} - F_{1})}^{2} + F_{2}^{2}}

3

\sqrt{{(F_{3} - F_{1})}^{2} - F_{2}^{2}}

4

F_{3} - F_{1} + F_{2}

Explanation:

B Since, it is placed in a uniform magnetic field.
Therefore, the total force on the current carrying in closed loop will be 0 .

Net horizontal force -
$F_{H} = F_{3} - F_{1}$
Net vertical force -
$F_{v} = F_{2}$
$∴$ Force on the segment QP is-
$| \vec{F} | = \sqrt{{(F_{3} - F_{1})}^{2} + F_{2}^{2}}$

CBESE AIPMT 2008

Moving Charges & Magnetism

153847 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $F$ is in plane of the coil. If due to a 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})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

3

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

C The current flowing clockwise in an equilateral triangle has a magnetic field in the direction of $\hat{k}$ .
$τ = BiNA \sin θ$
$τ = BiNA \sin 90^{\circ}$
$∴ τ = Bi \times \frac{\sqrt{3}}{4} l^{2} \times 1$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 {(\frac{τ}{Bi \sqrt{3}})}^{\frac{1}{2}}$

CBESE AIPMT 2005

Moving Charges & Magnetism

153844 A current carrying closed loop in the form of a right angled isosceles $△ ABC$ is placed in a uniform magnetic field acting along $AB$ . If the magnetic force on the $arm B C$ is $F$ , the force on the arm $AC$ is

1

- F

2

F

3

\sqrt{2 F}

4

- \sqrt{2 F}

Explanation:

A Force along $AB$ is zero as magnetic field is along $AB$ . The net force on a current carrying loop is always zero.
$F_{net} = F_{AB} + F_{BC} + F_{AC}$
$F_{AB} = 0 and F_{BC} = F$
Then, $0 = 0 + F + F_{AC}$
$F_{AC} = - F$

CBESE AIPMT 2011

Moving Charges & Magnetism

153845 A square current carrying loop is suspended in a uniform magnetic field acting in the place of the loop. If the force on one arm of the loop is $F$ , the net force on the remaining three arms of the loop is

1

3 F

2

- F

3

- 3 F

4

F

Explanation:

B :Consider a square PQRS which is suspended in a uniform magnetic field $B$ and force on one arm of the loop is $F$ .

As shown from the figure, side QR and PS are parallel to the direction of magnetic field. The force on these arm two arms of parallel to the field is zero.
$\vec{F} = i f (\vec{d} l \times \vec{B})$
$\vec{d} l =$ displacement and $\vec{B} =$ magnetic field
Hence, $F_{QR} = F_{PS} = 0$
Net force acting on the close loop
$F_{PQ} + F_{QR} + F_{RS} + F_{PS} = 0$
$F_{PQ} + 0 + F_{RS} + 0 = 0$
$F_{PQ} = - F_{RS}$
$F = - F_{RS}$
$F_{RS} = - F$

CBESE AIPMT 2010

Moving Charges & Magnetism

153846 A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are $F_{1}, F_{2}$ and $F_{3}$ respectively and are in the plane of the paper and along the directions shown in figure, the force on the segment $Q P$ is

1

F_{3} - F_{1} - F_{2}

2

\sqrt{{(F_{3} - F_{1})}^{2} + F_{2}^{2}}

3

\sqrt{{(F_{3} - F_{1})}^{2} - F_{2}^{2}}

4

F_{3} - F_{1} + F_{2}

Explanation:

B Since, it is placed in a uniform magnetic field.
Therefore, the total force on the current carrying in closed loop will be 0 .

Net horizontal force -
$F_{H} = F_{3} - F_{1}$
Net vertical force -
$F_{v} = F_{2}$
$∴$ Force on the segment QP is-
$| \vec{F} | = \sqrt{{(F_{3} - F_{1})}^{2} + F_{2}^{2}}$

CBESE AIPMT 2008

Moving Charges & Magnetism

153847 A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $F$ is in plane of the coil. If due to a 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})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{Bi})

3

2 {(\frac{τ}{\sqrt{3} Bi})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{Bi}

Explanation:

C The current flowing clockwise in an equilateral triangle has a magnetic field in the direction of $\hat{k}$ .
$τ = BiNA \sin θ$
$τ = BiNA \sin 90^{\circ}$
$∴ τ = Bi \times \frac{\sqrt{3}}{4} l^{2} \times 1$
$l^{2} = \frac{4 τ}{\sqrt{3} Bi}$
$l = 2 {(\frac{τ}{Bi \sqrt{3}})}^{\frac{1}{2}}$

CBESE AIPMT 2005