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Moving Charges & Magnetism

153123 A uniform magnetic field $\vec{B}$ is perpendicular to the plane of a circular loop of diameter $10 cm$ formed wire of diameter $2 mm$ and resistivity $2 \times 10^{- 8} Ω m$ . If a current of $11 A$ is to be induced in the loop then the rate at which $\vec{B}$ is to be changed is

1

2.8 {Ts}^{- 1}

2

1.4 {Ts}^{- 1}

3

3.2 {Ts}^{- 1}

4

2.4 {Ts}^{- 1}

Explanation:

A The resistance of the loop is
$R = ρ \frac{L}{A}$
$R = \frac{2 \times 10^{- 8} \times π \times 10 \times 10^{- 2}}{π {(2 \times 10^{- 3})}^{2 / 4}}$
$R = 20 \times 10^{- 2} Ω$
$I = | ε | / R$
$I \times R = | \frac{d ϕ_{B}}{dt} |$
$I \times R = A | \frac{dB}{dt} | [∴ ϕ_{B} = B . A]$
$| \frac{dB}{dt} | = \frac{IR}{π r^{2}} = \frac{11 \times 20 \times 10^{- 2}}{π \times {(5 \times 10^{- 2})}^{2}}$
$| \frac{dB}{dt} | = 2.8 {Ts}^{- 1}$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153124 A long current carrying wire produces a magnetic field of $1 T$ at a distance of $r$ . The magnetic field at (A) $\frac{r}{2}$ , (B) $2 r$ and (C) $3 r$ is

1

(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

2

(A) = 3 T, (B) = \frac{1}{3} T, (C) = \frac{1}{6} T

3

(A) = \frac{3}{2} T, (B) = \frac{1}{4} T, (C) = \frac{1}{8} T

4

(A) = \frac{5}{2} T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

Explanation:

A Given, $B = 1 T$
Magnetic field
Case-(A)
$B = \frac{μ_{o} I}{2 π r}$
Magnetic field at distance $\frac{r}{2}$ is
$B_{A} = \frac{μ_{o} I}{2 π (r / 2)} = 2 \times \frac{μ_{0} I}{2 π r} = 2 B = 2 \times 1 = 2 T$
Case-(B)
Magnetic field at distance (2r) is
Case-(C)
$B_{B} = \frac{μ_{o} I}{2 π \times 2 r} = \frac{1}{2} \times \frac{μ_{0} I}{2 π r} = \frac{1}{2} B = \frac{1}{2} \times 1 = \frac{1}{2} T$
Magnetic field at distance (3r) is
$B_{C} = \frac{μ_{o} I}{2 π \times 3 r} = \frac{1}{3} \times \frac{μ_{0} I}{2 π r} = \frac{1}{3} B = \frac{1}{3} \times 1 = \frac{1}{3} T$
From case (A) and (B) and (C)
$(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153125 A compass needle oscillates 20 times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_{1}$ . The same needle oscillates 30 times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_{2}$ . Then $B_{1} : B_{2}$ is

1

9 \sqrt{3} : 4 \sqrt{2}

2

4 \sqrt{2} : 9 \sqrt{3}

3

3 \sqrt{3} : 2 \sqrt{2}

4

2 \sqrt{2} : 3 \sqrt{3}

Explanation:

D $\frac{B_{H_{2}}}{B_{H_{1}}} = \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{B_{2} \cos θ_{2}}{B_{1} \cos θ_{1}} = \frac{T_{1}^{2}}{T_{2}^{2}}$
$\frac{B_{2}}{B_{1}} = \frac{T_{1}^{2} \cos θ_{1}}{T_{2}^{2} \cos θ_{2}}$
$= {(\frac{3}{2})}^{2} \frac{\cos 45^{\circ}}{\cos 30^{\circ}} = {(\frac{3}{2})}^{2}$
$\frac{B_{2}}{B_{1}} = \frac{3 \sqrt{3}}{2 \sqrt{2}}$
$\frac{B_{1}}{B_{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}} Or 2 \sqrt{2} : 3 \sqrt{3}$
Then $B_{1} : B_{2} = 2 \sqrt{2} : 3 \sqrt{3}$

AP EAMCET-04.07.2022

Moving Charges & Magnetism

153126 A rectangular coil of length $2 cm$ and width $1.25 cm$ with 250 turns carries a current of 55 $μ A$ and subjected to a magnetic field of strength 0.64 T. Work done for rotating the coil by $180^{\circ}$ against the torque is

1

2.2 μ J

2

3.5 μ J

3

4.4 μ J

4

5.5 μ J

Explanation:

C Given,
$l = 2 cm = 2 \times 10^{- 2} m$
$b = 1.25 cm = 1.25 \times 10^{- 2} m$
$N = 250$
$I = 55 μ A = 55 \times 10^{- 6} A$
$B = 0.64 T$
$θ = 180^{\circ}$
Work done in rotating the coil
$W = MB (1 - \cos θ)$
$W = MB (1 - \cos 180^{\circ})$
$W = 2 MB$
$W = 2 NIAB$
$W = 2 NI \times l \times b \times B$
$= 2 \times 250 \times 55 \times 10^{- 6} \times 2 \times 10^{- 2} \times 1.25 \times 10^{- 2} \times 0.64$
$W = 4.4 \times 10^{- 6} J$
$W = 4.4 μ J$
$W = 4.4 \times 10^{- 6} J$

AP EAMCET-12.07.2022

Moving Charges & Magnetism

153127 An isosceles triangular current carrying loop is kept perpendicularly in a uniform magnetic field ${\vec{B}}_{0}$ as shown in figure. The force on the loop due to magnetic field is

1

{ILB}_{0} \cos θ

2

2 {ILB}_{0} \cos θ

3 0

4 ILB

_{0} \sin θ

Explanation:

C
Resultant of $F_{PQ}, F_{QR}$ and $F_{RP}$ is zero, so net force on entire loop $PQR$ is zero.

AP EAMCET-05.07.2022

Moving Charges & Magnetism

153123 A uniform magnetic field $\vec{B}$ is perpendicular to the plane of a circular loop of diameter $10 cm$ formed wire of diameter $2 mm$ and resistivity $2 \times 10^{- 8} Ω m$ . If a current of $11 A$ is to be induced in the loop then the rate at which $\vec{B}$ is to be changed is

1

2.8 {Ts}^{- 1}

2

1.4 {Ts}^{- 1}

3

3.2 {Ts}^{- 1}

4

2.4 {Ts}^{- 1}

Explanation:

A The resistance of the loop is
$R = ρ \frac{L}{A}$
$R = \frac{2 \times 10^{- 8} \times π \times 10 \times 10^{- 2}}{π {(2 \times 10^{- 3})}^{2 / 4}}$
$R = 20 \times 10^{- 2} Ω$
$I = | ε | / R$
$I \times R = | \frac{d ϕ_{B}}{dt} |$
$I \times R = A | \frac{dB}{dt} | [∴ ϕ_{B} = B . A]$
$| \frac{dB}{dt} | = \frac{IR}{π r^{2}} = \frac{11 \times 20 \times 10^{- 2}}{π \times {(5 \times 10^{- 2})}^{2}}$
$| \frac{dB}{dt} | = 2.8 {Ts}^{- 1}$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153124 A long current carrying wire produces a magnetic field of $1 T$ at a distance of $r$ . The magnetic field at (A) $\frac{r}{2}$ , (B) $2 r$ and (C) $3 r$ is

1

(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

2

(A) = 3 T, (B) = \frac{1}{3} T, (C) = \frac{1}{6} T

3

(A) = \frac{3}{2} T, (B) = \frac{1}{4} T, (C) = \frac{1}{8} T

4

(A) = \frac{5}{2} T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

Explanation:

A Given, $B = 1 T$
Magnetic field
Case-(A)
$B = \frac{μ_{o} I}{2 π r}$
Magnetic field at distance $\frac{r}{2}$ is
$B_{A} = \frac{μ_{o} I}{2 π (r / 2)} = 2 \times \frac{μ_{0} I}{2 π r} = 2 B = 2 \times 1 = 2 T$
Case-(B)
Magnetic field at distance (2r) is
Case-(C)
$B_{B} = \frac{μ_{o} I}{2 π \times 2 r} = \frac{1}{2} \times \frac{μ_{0} I}{2 π r} = \frac{1}{2} B = \frac{1}{2} \times 1 = \frac{1}{2} T$
Magnetic field at distance (3r) is
$B_{C} = \frac{μ_{o} I}{2 π \times 3 r} = \frac{1}{3} \times \frac{μ_{0} I}{2 π r} = \frac{1}{3} B = \frac{1}{3} \times 1 = \frac{1}{3} T$
From case (A) and (B) and (C)
$(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153125 A compass needle oscillates 20 times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_{1}$ . The same needle oscillates 30 times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_{2}$ . Then $B_{1} : B_{2}$ is

1

9 \sqrt{3} : 4 \sqrt{2}

2

4 \sqrt{2} : 9 \sqrt{3}

3

3 \sqrt{3} : 2 \sqrt{2}

4

2 \sqrt{2} : 3 \sqrt{3}

Explanation:

D $\frac{B_{H_{2}}}{B_{H_{1}}} = \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{B_{2} \cos θ_{2}}{B_{1} \cos θ_{1}} = \frac{T_{1}^{2}}{T_{2}^{2}}$
$\frac{B_{2}}{B_{1}} = \frac{T_{1}^{2} \cos θ_{1}}{T_{2}^{2} \cos θ_{2}}$
$= {(\frac{3}{2})}^{2} \frac{\cos 45^{\circ}}{\cos 30^{\circ}} = {(\frac{3}{2})}^{2}$
$\frac{B_{2}}{B_{1}} = \frac{3 \sqrt{3}}{2 \sqrt{2}}$
$\frac{B_{1}}{B_{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}} Or 2 \sqrt{2} : 3 \sqrt{3}$
Then $B_{1} : B_{2} = 2 \sqrt{2} : 3 \sqrt{3}$

AP EAMCET-04.07.2022

Moving Charges & Magnetism

153126 A rectangular coil of length $2 cm$ and width $1.25 cm$ with 250 turns carries a current of 55 $μ A$ and subjected to a magnetic field of strength 0.64 T. Work done for rotating the coil by $180^{\circ}$ against the torque is

1

2.2 μ J

2

3.5 μ J

3

4.4 μ J

4

5.5 μ J

Explanation:

C Given,
$l = 2 cm = 2 \times 10^{- 2} m$
$b = 1.25 cm = 1.25 \times 10^{- 2} m$
$N = 250$
$I = 55 μ A = 55 \times 10^{- 6} A$
$B = 0.64 T$
$θ = 180^{\circ}$
Work done in rotating the coil
$W = MB (1 - \cos θ)$
$W = MB (1 - \cos 180^{\circ})$
$W = 2 MB$
$W = 2 NIAB$
$W = 2 NI \times l \times b \times B$
$= 2 \times 250 \times 55 \times 10^{- 6} \times 2 \times 10^{- 2} \times 1.25 \times 10^{- 2} \times 0.64$
$W = 4.4 \times 10^{- 6} J$
$W = 4.4 μ J$
$W = 4.4 \times 10^{- 6} J$

AP EAMCET-12.07.2022

Moving Charges & Magnetism

153127 An isosceles triangular current carrying loop is kept perpendicularly in a uniform magnetic field ${\vec{B}}_{0}$ as shown in figure. The force on the loop due to magnetic field is

1

{ILB}_{0} \cos θ

2

2 {ILB}_{0} \cos θ

3 0

4 ILB

_{0} \sin θ

Explanation:

C
Resultant of $F_{PQ}, F_{QR}$ and $F_{RP}$ is zero, so net force on entire loop $PQR$ is zero.

AP EAMCET-05.07.2022

Moving Charges & Magnetism

153123 A uniform magnetic field $\vec{B}$ is perpendicular to the plane of a circular loop of diameter $10 cm$ formed wire of diameter $2 mm$ and resistivity $2 \times 10^{- 8} Ω m$ . If a current of $11 A$ is to be induced in the loop then the rate at which $\vec{B}$ is to be changed is

1

2.8 {Ts}^{- 1}

2

1.4 {Ts}^{- 1}

3

3.2 {Ts}^{- 1}

4

2.4 {Ts}^{- 1}

Explanation:

A The resistance of the loop is
$R = ρ \frac{L}{A}$
$R = \frac{2 \times 10^{- 8} \times π \times 10 \times 10^{- 2}}{π {(2 \times 10^{- 3})}^{2 / 4}}$
$R = 20 \times 10^{- 2} Ω$
$I = | ε | / R$
$I \times R = | \frac{d ϕ_{B}}{dt} |$
$I \times R = A | \frac{dB}{dt} | [∴ ϕ_{B} = B . A]$
$| \frac{dB}{dt} | = \frac{IR}{π r^{2}} = \frac{11 \times 20 \times 10^{- 2}}{π \times {(5 \times 10^{- 2})}^{2}}$
$| \frac{dB}{dt} | = 2.8 {Ts}^{- 1}$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153124 A long current carrying wire produces a magnetic field of $1 T$ at a distance of $r$ . The magnetic field at (A) $\frac{r}{2}$ , (B) $2 r$ and (C) $3 r$ is

1

(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

2

(A) = 3 T, (B) = \frac{1}{3} T, (C) = \frac{1}{6} T

3

(A) = \frac{3}{2} T, (B) = \frac{1}{4} T, (C) = \frac{1}{8} T

4

(A) = \frac{5}{2} T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

Explanation:

A Given, $B = 1 T$
Magnetic field
Case-(A)
$B = \frac{μ_{o} I}{2 π r}$
Magnetic field at distance $\frac{r}{2}$ is
$B_{A} = \frac{μ_{o} I}{2 π (r / 2)} = 2 \times \frac{μ_{0} I}{2 π r} = 2 B = 2 \times 1 = 2 T$
Case-(B)
Magnetic field at distance (2r) is
Case-(C)
$B_{B} = \frac{μ_{o} I}{2 π \times 2 r} = \frac{1}{2} \times \frac{μ_{0} I}{2 π r} = \frac{1}{2} B = \frac{1}{2} \times 1 = \frac{1}{2} T$
Magnetic field at distance (3r) is
$B_{C} = \frac{μ_{o} I}{2 π \times 3 r} = \frac{1}{3} \times \frac{μ_{0} I}{2 π r} = \frac{1}{3} B = \frac{1}{3} \times 1 = \frac{1}{3} T$
From case (A) and (B) and (C)
$(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153125 A compass needle oscillates 20 times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_{1}$ . The same needle oscillates 30 times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_{2}$ . Then $B_{1} : B_{2}$ is

1

9 \sqrt{3} : 4 \sqrt{2}

2

4 \sqrt{2} : 9 \sqrt{3}

3

3 \sqrt{3} : 2 \sqrt{2}

4

2 \sqrt{2} : 3 \sqrt{3}

Explanation:

D $\frac{B_{H_{2}}}{B_{H_{1}}} = \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{B_{2} \cos θ_{2}}{B_{1} \cos θ_{1}} = \frac{T_{1}^{2}}{T_{2}^{2}}$
$\frac{B_{2}}{B_{1}} = \frac{T_{1}^{2} \cos θ_{1}}{T_{2}^{2} \cos θ_{2}}$
$= {(\frac{3}{2})}^{2} \frac{\cos 45^{\circ}}{\cos 30^{\circ}} = {(\frac{3}{2})}^{2}$
$\frac{B_{2}}{B_{1}} = \frac{3 \sqrt{3}}{2 \sqrt{2}}$
$\frac{B_{1}}{B_{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}} Or 2 \sqrt{2} : 3 \sqrt{3}$
Then $B_{1} : B_{2} = 2 \sqrt{2} : 3 \sqrt{3}$

AP EAMCET-04.07.2022

Moving Charges & Magnetism

153126 A rectangular coil of length $2 cm$ and width $1.25 cm$ with 250 turns carries a current of 55 $μ A$ and subjected to a magnetic field of strength 0.64 T. Work done for rotating the coil by $180^{\circ}$ against the torque is

1

2.2 μ J

2

3.5 μ J

3

4.4 μ J

4

5.5 μ J

Explanation:

C Given,
$l = 2 cm = 2 \times 10^{- 2} m$
$b = 1.25 cm = 1.25 \times 10^{- 2} m$
$N = 250$
$I = 55 μ A = 55 \times 10^{- 6} A$
$B = 0.64 T$
$θ = 180^{\circ}$
Work done in rotating the coil
$W = MB (1 - \cos θ)$
$W = MB (1 - \cos 180^{\circ})$
$W = 2 MB$
$W = 2 NIAB$
$W = 2 NI \times l \times b \times B$
$= 2 \times 250 \times 55 \times 10^{- 6} \times 2 \times 10^{- 2} \times 1.25 \times 10^{- 2} \times 0.64$
$W = 4.4 \times 10^{- 6} J$
$W = 4.4 μ J$
$W = 4.4 \times 10^{- 6} J$

AP EAMCET-12.07.2022

Moving Charges & Magnetism

153127 An isosceles triangular current carrying loop is kept perpendicularly in a uniform magnetic field ${\vec{B}}_{0}$ as shown in figure. The force on the loop due to magnetic field is

1

{ILB}_{0} \cos θ

2

2 {ILB}_{0} \cos θ

3 0

4 ILB

_{0} \sin θ

Explanation:

C
Resultant of $F_{PQ}, F_{QR}$ and $F_{RP}$ is zero, so net force on entire loop $PQR$ is zero.

AP EAMCET-05.07.2022

Moving Charges & Magnetism

153123 A uniform magnetic field $\vec{B}$ is perpendicular to the plane of a circular loop of diameter $10 cm$ formed wire of diameter $2 mm$ and resistivity $2 \times 10^{- 8} Ω m$ . If a current of $11 A$ is to be induced in the loop then the rate at which $\vec{B}$ is to be changed is

1

2.8 {Ts}^{- 1}

2

1.4 {Ts}^{- 1}

3

3.2 {Ts}^{- 1}

4

2.4 {Ts}^{- 1}

Explanation:

A The resistance of the loop is
$R = ρ \frac{L}{A}$
$R = \frac{2 \times 10^{- 8} \times π \times 10 \times 10^{- 2}}{π {(2 \times 10^{- 3})}^{2 / 4}}$
$R = 20 \times 10^{- 2} Ω$
$I = | ε | / R$
$I \times R = | \frac{d ϕ_{B}}{dt} |$
$I \times R = A | \frac{dB}{dt} | [∴ ϕ_{B} = B . A]$
$| \frac{dB}{dt} | = \frac{IR}{π r^{2}} = \frac{11 \times 20 \times 10^{- 2}}{π \times {(5 \times 10^{- 2})}^{2}}$
$| \frac{dB}{dt} | = 2.8 {Ts}^{- 1}$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153124 A long current carrying wire produces a magnetic field of $1 T$ at a distance of $r$ . The magnetic field at (A) $\frac{r}{2}$ , (B) $2 r$ and (C) $3 r$ is

1

(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

2

(A) = 3 T, (B) = \frac{1}{3} T, (C) = \frac{1}{6} T

3

(A) = \frac{3}{2} T, (B) = \frac{1}{4} T, (C) = \frac{1}{8} T

4

(A) = \frac{5}{2} T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

Explanation:

A Given, $B = 1 T$
Magnetic field
Case-(A)
$B = \frac{μ_{o} I}{2 π r}$
Magnetic field at distance $\frac{r}{2}$ is
$B_{A} = \frac{μ_{o} I}{2 π (r / 2)} = 2 \times \frac{μ_{0} I}{2 π r} = 2 B = 2 \times 1 = 2 T$
Case-(B)
Magnetic field at distance (2r) is
Case-(C)
$B_{B} = \frac{μ_{o} I}{2 π \times 2 r} = \frac{1}{2} \times \frac{μ_{0} I}{2 π r} = \frac{1}{2} B = \frac{1}{2} \times 1 = \frac{1}{2} T$
Magnetic field at distance (3r) is
$B_{C} = \frac{μ_{o} I}{2 π \times 3 r} = \frac{1}{3} \times \frac{μ_{0} I}{2 π r} = \frac{1}{3} B = \frac{1}{3} \times 1 = \frac{1}{3} T$
From case (A) and (B) and (C)
$(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153125 A compass needle oscillates 20 times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_{1}$ . The same needle oscillates 30 times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_{2}$ . Then $B_{1} : B_{2}$ is

1

9 \sqrt{3} : 4 \sqrt{2}

2

4 \sqrt{2} : 9 \sqrt{3}

3

3 \sqrt{3} : 2 \sqrt{2}

4

2 \sqrt{2} : 3 \sqrt{3}

Explanation:

D $\frac{B_{H_{2}}}{B_{H_{1}}} = \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{B_{2} \cos θ_{2}}{B_{1} \cos θ_{1}} = \frac{T_{1}^{2}}{T_{2}^{2}}$
$\frac{B_{2}}{B_{1}} = \frac{T_{1}^{2} \cos θ_{1}}{T_{2}^{2} \cos θ_{2}}$
$= {(\frac{3}{2})}^{2} \frac{\cos 45^{\circ}}{\cos 30^{\circ}} = {(\frac{3}{2})}^{2}$
$\frac{B_{2}}{B_{1}} = \frac{3 \sqrt{3}}{2 \sqrt{2}}$
$\frac{B_{1}}{B_{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}} Or 2 \sqrt{2} : 3 \sqrt{3}$
Then $B_{1} : B_{2} = 2 \sqrt{2} : 3 \sqrt{3}$

AP EAMCET-04.07.2022

Moving Charges & Magnetism

153126 A rectangular coil of length $2 cm$ and width $1.25 cm$ with 250 turns carries a current of 55 $μ A$ and subjected to a magnetic field of strength 0.64 T. Work done for rotating the coil by $180^{\circ}$ against the torque is

1

2.2 μ J

2

3.5 μ J

3

4.4 μ J

4

5.5 μ J

Explanation:

C Given,
$l = 2 cm = 2 \times 10^{- 2} m$
$b = 1.25 cm = 1.25 \times 10^{- 2} m$
$N = 250$
$I = 55 μ A = 55 \times 10^{- 6} A$
$B = 0.64 T$
$θ = 180^{\circ}$
Work done in rotating the coil
$W = MB (1 - \cos θ)$
$W = MB (1 - \cos 180^{\circ})$
$W = 2 MB$
$W = 2 NIAB$
$W = 2 NI \times l \times b \times B$
$= 2 \times 250 \times 55 \times 10^{- 6} \times 2 \times 10^{- 2} \times 1.25 \times 10^{- 2} \times 0.64$
$W = 4.4 \times 10^{- 6} J$
$W = 4.4 μ J$
$W = 4.4 \times 10^{- 6} J$

AP EAMCET-12.07.2022

Moving Charges & Magnetism

153127 An isosceles triangular current carrying loop is kept perpendicularly in a uniform magnetic field ${\vec{B}}_{0}$ as shown in figure. The force on the loop due to magnetic field is

1

{ILB}_{0} \cos θ

2

2 {ILB}_{0} \cos θ

3 0

4 ILB

_{0} \sin θ

Explanation:

C
Resultant of $F_{PQ}, F_{QR}$ and $F_{RP}$ is zero, so net force on entire loop $PQR$ is zero.

AP EAMCET-05.07.2022

Moving Charges & Magnetism

153123 A uniform magnetic field $\vec{B}$ is perpendicular to the plane of a circular loop of diameter $10 cm$ formed wire of diameter $2 mm$ and resistivity $2 \times 10^{- 8} Ω m$ . If a current of $11 A$ is to be induced in the loop then the rate at which $\vec{B}$ is to be changed is

1

2.8 {Ts}^{- 1}

2

1.4 {Ts}^{- 1}

3

3.2 {Ts}^{- 1}

4

2.4 {Ts}^{- 1}

Explanation:

A The resistance of the loop is
$R = ρ \frac{L}{A}$
$R = \frac{2 \times 10^{- 8} \times π \times 10 \times 10^{- 2}}{π {(2 \times 10^{- 3})}^{2 / 4}}$
$R = 20 \times 10^{- 2} Ω$
$I = | ε | / R$
$I \times R = | \frac{d ϕ_{B}}{dt} |$
$I \times R = A | \frac{dB}{dt} | [∴ ϕ_{B} = B . A]$
$| \frac{dB}{dt} | = \frac{IR}{π r^{2}} = \frac{11 \times 20 \times 10^{- 2}}{π \times {(5 \times 10^{- 2})}^{2}}$
$| \frac{dB}{dt} | = 2.8 {Ts}^{- 1}$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153124 A long current carrying wire produces a magnetic field of $1 T$ at a distance of $r$ . The magnetic field at (A) $\frac{r}{2}$ , (B) $2 r$ and (C) $3 r$ is

1

(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

2

(A) = 3 T, (B) = \frac{1}{3} T, (C) = \frac{1}{6} T

3

(A) = \frac{3}{2} T, (B) = \frac{1}{4} T, (C) = \frac{1}{8} T

4

(A) = \frac{5}{2} T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T

Explanation:

A Given, $B = 1 T$
Magnetic field
Case-(A)
$B = \frac{μ_{o} I}{2 π r}$
Magnetic field at distance $\frac{r}{2}$ is
$B_{A} = \frac{μ_{o} I}{2 π (r / 2)} = 2 \times \frac{μ_{0} I}{2 π r} = 2 B = 2 \times 1 = 2 T$
Case-(B)
Magnetic field at distance (2r) is
Case-(C)
$B_{B} = \frac{μ_{o} I}{2 π \times 2 r} = \frac{1}{2} \times \frac{μ_{0} I}{2 π r} = \frac{1}{2} B = \frac{1}{2} \times 1 = \frac{1}{2} T$
Magnetic field at distance (3r) is
$B_{C} = \frac{μ_{o} I}{2 π \times 3 r} = \frac{1}{3} \times \frac{μ_{0} I}{2 π r} = \frac{1}{3} B = \frac{1}{3} \times 1 = \frac{1}{3} T$
From case (A) and (B) and (C)
$(A) = 2 T, (B) = \frac{1}{2} T, (C) = \frac{1}{3} T$

AP EAMCET-06.07.2022

Moving Charges & Magnetism

153125 A compass needle oscillates 20 times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_{1}$ . The same needle oscillates 30 times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_{2}$ . Then $B_{1} : B_{2}$ is

1

9 \sqrt{3} : 4 \sqrt{2}

2

4 \sqrt{2} : 9 \sqrt{3}

3

3 \sqrt{3} : 2 \sqrt{2}

4

2 \sqrt{2} : 3 \sqrt{3}

Explanation:

D $\frac{B_{H_{2}}}{B_{H_{1}}} = \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{B_{2} \cos θ_{2}}{B_{1} \cos θ_{1}} = \frac{T_{1}^{2}}{T_{2}^{2}}$
$\frac{B_{2}}{B_{1}} = \frac{T_{1}^{2} \cos θ_{1}}{T_{2}^{2} \cos θ_{2}}$
$= {(\frac{3}{2})}^{2} \frac{\cos 45^{\circ}}{\cos 30^{\circ}} = {(\frac{3}{2})}^{2}$
$\frac{B_{2}}{B_{1}} = \frac{3 \sqrt{3}}{2 \sqrt{2}}$
$\frac{B_{1}}{B_{2}} = \frac{2 \sqrt{2}}{3 \sqrt{3}} Or 2 \sqrt{2} : 3 \sqrt{3}$
Then $B_{1} : B_{2} = 2 \sqrt{2} : 3 \sqrt{3}$

AP EAMCET-04.07.2022

Moving Charges & Magnetism

153126 A rectangular coil of length $2 cm$ and width $1.25 cm$ with 250 turns carries a current of 55 $μ A$ and subjected to a magnetic field of strength 0.64 T. Work done for rotating the coil by $180^{\circ}$ against the torque is

1

2.2 μ J

2

3.5 μ J

3

4.4 μ J

4

5.5 μ J

Explanation:

C Given,
$l = 2 cm = 2 \times 10^{- 2} m$
$b = 1.25 cm = 1.25 \times 10^{- 2} m$
$N = 250$
$I = 55 μ A = 55 \times 10^{- 6} A$
$B = 0.64 T$
$θ = 180^{\circ}$
Work done in rotating the coil
$W = MB (1 - \cos θ)$
$W = MB (1 - \cos 180^{\circ})$
$W = 2 MB$
$W = 2 NIAB$
$W = 2 NI \times l \times b \times B$
$= 2 \times 250 \times 55 \times 10^{- 6} \times 2 \times 10^{- 2} \times 1.25 \times 10^{- 2} \times 0.64$
$W = 4.4 \times 10^{- 6} J$
$W = 4.4 μ J$
$W = 4.4 \times 10^{- 6} J$

AP EAMCET-12.07.2022

Moving Charges & Magnetism

153127 An isosceles triangular current carrying loop is kept perpendicularly in a uniform magnetic field ${\vec{B}}_{0}$ as shown in figure. The force on the loop due to magnetic field is

1

{ILB}_{0} \cos θ

2

2 {ILB}_{0} \cos θ

3 0

4 ILB

_{0} \sin θ

Explanation:

C
Resultant of $F_{PQ}, F_{QR}$ and $F_{RP}$ is zero, so net force on entire loop $PQR$ is zero.

AP EAMCET-05.07.2022