QBManager

Moving Charges & Magnetism

153924 Two short bar magnets $A$ and $B$ are arranged coaxially. The distance between their centers is $30$ cm. A compass needle placed on their axis at a distance of $6 cm$ from $B$ shows no deflection. The ratio of the magnetic moments of $A$ and $B$ is

1

16 : 1

2

1 : 16

3

64 : 1

4

1 : 64

Explanation:

C Distance between the magnets (d) $= 30 cm$ Compass needle distance from magnet $(B) = 6 cm$ Now, distance of magnet $A$ from needle, $d_{1} = 30 - 6 = 24 cm$
and distance of magnet $B$ from needle $(d_{2}) = 6 cm$
$∴$ Magnetic field on a axial line is given as,
$B = \frac{μ_{o}}{4 π} \cdot \frac{2 M}{d^{3}}$
Hence, the ratio of the magnetic moment for $A$ and $B$
$\frac{M_{1}}{d_{1}^{3}} = \frac{M_{2}}{d_{2}^{3}}$
$\frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{24}{6})}^{3}$
$\frac{M_{1}}{M_{2}} = 4^{3} = 64$
So, the ratio of the magnetic moment of $A$ and $B$ is $64 : 1$

AP EAMCET (22.04.2019) Shift-II

Moving Charges & Magnetism

153925 A regular hexagon of side a. A wire of length 24a is coiled on that hexagon. If current in hexagon is $I$ , then find the magnetic moment.

1

6 \sqrt{3} {Ia}^{2}

2

3 \sqrt{3} {Ia}^{2}

3

\frac{3 \sqrt{3}}{2} {Is}^{2}

4

6 {Ia}^{2}

Explanation:

A Length of wire $= 24 a$
Side of hexagon $= a$
Total length of hexagon (circumference) $= 6 a$
Let, number of turns in hexagon $= n$
$24 a = n \times 6 a$
$n = 4$
Area of hexagon $(A) = \frac{3 \sqrt{3}}{2} x^{2} (x =$ side of hexagon $)$
$A = \frac{3 \sqrt{3} a^{2}}{2}$
$∴$ Magnetic moment $(M) = nIA$
$M = 4 \times I \times \frac{3 \sqrt{3} a^{2}}{2}$
$M = 6 \sqrt{3} I a^{2}$

JIPMER-2018

Moving Charges & Magnetism

153926 A moving coil galvanometer has a rectangular wire coil of enclosed area $0.001 m^{2}$ and 500 turns. The coil operates in a radial magnetic field of $0.2 T$ and carries a current of $6 π \times 10^{- 8} A$ . If the torsional spring constant is $6 \times 10^{- 7} N - m / rad$ , then the angular deflection of the coil in radians is

1

\frac{π}{100}

2

\frac{π}{200}

3

\frac{π}{300}

4

\frac{π}{400}

Explanation:

A Given that, $n = 500$ turns, $B = 0.2 T$
Area $(A) = 0.001 m^{2}, I = 6 π \times 10^{- 8} A$
$k = 6 \times 10^{- 7} N - m / rad$
We know that, the expansion of angular deflection of the coil $(θ) = (\frac{nBA}{k}) I$
$θ = \frac{500 \times 0.2 \times 0.001}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = \frac{5 \times 10^{2} \times 2 \times 10^{- 4}}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = π \times 10^{3} \times 10^{7} \times 10^{- 12} = π \times 10^{- 2}$
$θ = \frac{π}{100}$

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153927 The magnetic field in a travelling electromagnetic wave has a peak value of 20nT. The peak value of electric field strength is

1

3 V / m

2

6 V / m

3

9 V / m

4

12 V / m

Explanation:

B Given that,
$B_{0} = 20 nT = 20 \times 10^{- 9}$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 20 \times 10^{- 9}$
$E_{o} = \frac{6 V}{m}$

VITEEE-2018

Moving Charges & Magnetism

153928 A coil in the shape of an equilateral triangle of side $l$ is suspended between two pole pieces of a permanent magnet, such that magnetic field, $B$ 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

2 {(\frac{τ}{\sqrt{3} BI})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{BI})

3

2 {(\frac{τ}{BI})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{BI}

Explanation:

A
Magnetic moment $(\vec{M}) = \vec{I} \times \vec{A}$
Torque $(\vec{τ}) = \vec{M} \times \vec{B}$
$τ = MB \sin θ$
$(θ = 90^{\circ})$
$τ = MB \sin 90^{\circ}$
$τ = MB$
$τ = IAB$
$(∵ M = IA)$
Area of equilateral triangle of side length $= \frac{\sqrt{3}}{4} l^{2}$
$\frac{\sqrt{3}}{4} l^{2} = \frac{τ}{IB}$
$l^{2} = \frac{4 τ}{\sqrt{3} IB}$
$l^{2} = 2 {(\frac{τ}{\sqrt{3} IB})}^{1 / 2}$

BCECE-2017

Moving Charges & Magnetism

153924 Two short bar magnets $A$ and $B$ are arranged coaxially. The distance between their centers is $30$ cm. A compass needle placed on their axis at a distance of $6 cm$ from $B$ shows no deflection. The ratio of the magnetic moments of $A$ and $B$ is

1

16 : 1

2

1 : 16

3

64 : 1

4

1 : 64

Explanation:

C Distance between the magnets (d) $= 30 cm$ Compass needle distance from magnet $(B) = 6 cm$ Now, distance of magnet $A$ from needle, $d_{1} = 30 - 6 = 24 cm$
and distance of magnet $B$ from needle $(d_{2}) = 6 cm$
$∴$ Magnetic field on a axial line is given as,
$B = \frac{μ_{o}}{4 π} \cdot \frac{2 M}{d^{3}}$
Hence, the ratio of the magnetic moment for $A$ and $B$
$\frac{M_{1}}{d_{1}^{3}} = \frac{M_{2}}{d_{2}^{3}}$
$\frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{24}{6})}^{3}$
$\frac{M_{1}}{M_{2}} = 4^{3} = 64$
So, the ratio of the magnetic moment of $A$ and $B$ is $64 : 1$

AP EAMCET (22.04.2019) Shift-II

Moving Charges & Magnetism

153925 A regular hexagon of side a. A wire of length 24a is coiled on that hexagon. If current in hexagon is $I$ , then find the magnetic moment.

1

6 \sqrt{3} {Ia}^{2}

2

3 \sqrt{3} {Ia}^{2}

3

\frac{3 \sqrt{3}}{2} {Is}^{2}

4

6 {Ia}^{2}

Explanation:

A Length of wire $= 24 a$
Side of hexagon $= a$
Total length of hexagon (circumference) $= 6 a$
Let, number of turns in hexagon $= n$
$24 a = n \times 6 a$
$n = 4$
Area of hexagon $(A) = \frac{3 \sqrt{3}}{2} x^{2} (x =$ side of hexagon $)$
$A = \frac{3 \sqrt{3} a^{2}}{2}$
$∴$ Magnetic moment $(M) = nIA$
$M = 4 \times I \times \frac{3 \sqrt{3} a^{2}}{2}$
$M = 6 \sqrt{3} I a^{2}$

JIPMER-2018

Moving Charges & Magnetism

153926 A moving coil galvanometer has a rectangular wire coil of enclosed area $0.001 m^{2}$ and 500 turns. The coil operates in a radial magnetic field of $0.2 T$ and carries a current of $6 π \times 10^{- 8} A$ . If the torsional spring constant is $6 \times 10^{- 7} N - m / rad$ , then the angular deflection of the coil in radians is

1

\frac{π}{100}

2

\frac{π}{200}

3

\frac{π}{300}

4

\frac{π}{400}

Explanation:

A Given that, $n = 500$ turns, $B = 0.2 T$
Area $(A) = 0.001 m^{2}, I = 6 π \times 10^{- 8} A$
$k = 6 \times 10^{- 7} N - m / rad$
We know that, the expansion of angular deflection of the coil $(θ) = (\frac{nBA}{k}) I$
$θ = \frac{500 \times 0.2 \times 0.001}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = \frac{5 \times 10^{2} \times 2 \times 10^{- 4}}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = π \times 10^{3} \times 10^{7} \times 10^{- 12} = π \times 10^{- 2}$
$θ = \frac{π}{100}$

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153927 The magnetic field in a travelling electromagnetic wave has a peak value of 20nT. The peak value of electric field strength is

1

3 V / m

2

6 V / m

3

9 V / m

4

12 V / m

Explanation:

B Given that,
$B_{0} = 20 nT = 20 \times 10^{- 9}$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 20 \times 10^{- 9}$
$E_{o} = \frac{6 V}{m}$

VITEEE-2018

Moving Charges & Magnetism

153928 A coil in the shape of an equilateral triangle of side $l$ is suspended between two pole pieces of a permanent magnet, such that magnetic field, $B$ 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

2 {(\frac{τ}{\sqrt{3} BI})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{BI})

3

2 {(\frac{τ}{BI})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{BI}

Explanation:

A
Magnetic moment $(\vec{M}) = \vec{I} \times \vec{A}$
Torque $(\vec{τ}) = \vec{M} \times \vec{B}$
$τ = MB \sin θ$
$(θ = 90^{\circ})$
$τ = MB \sin 90^{\circ}$
$τ = MB$
$τ = IAB$
$(∵ M = IA)$
Area of equilateral triangle of side length $= \frac{\sqrt{3}}{4} l^{2}$
$\frac{\sqrt{3}}{4} l^{2} = \frac{τ}{IB}$
$l^{2} = \frac{4 τ}{\sqrt{3} IB}$
$l^{2} = 2 {(\frac{τ}{\sqrt{3} IB})}^{1 / 2}$

BCECE-2017

Moving Charges & Magnetism

153924 Two short bar magnets $A$ and $B$ are arranged coaxially. The distance between their centers is $30$ cm. A compass needle placed on their axis at a distance of $6 cm$ from $B$ shows no deflection. The ratio of the magnetic moments of $A$ and $B$ is

1

16 : 1

2

1 : 16

3

64 : 1

4

1 : 64

Explanation:

C Distance between the magnets (d) $= 30 cm$ Compass needle distance from magnet $(B) = 6 cm$ Now, distance of magnet $A$ from needle, $d_{1} = 30 - 6 = 24 cm$
and distance of magnet $B$ from needle $(d_{2}) = 6 cm$
$∴$ Magnetic field on a axial line is given as,
$B = \frac{μ_{o}}{4 π} \cdot \frac{2 M}{d^{3}}$
Hence, the ratio of the magnetic moment for $A$ and $B$
$\frac{M_{1}}{d_{1}^{3}} = \frac{M_{2}}{d_{2}^{3}}$
$\frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{24}{6})}^{3}$
$\frac{M_{1}}{M_{2}} = 4^{3} = 64$
So, the ratio of the magnetic moment of $A$ and $B$ is $64 : 1$

AP EAMCET (22.04.2019) Shift-II

Moving Charges & Magnetism

153925 A regular hexagon of side a. A wire of length 24a is coiled on that hexagon. If current in hexagon is $I$ , then find the magnetic moment.

1

6 \sqrt{3} {Ia}^{2}

2

3 \sqrt{3} {Ia}^{2}

3

\frac{3 \sqrt{3}}{2} {Is}^{2}

4

6 {Ia}^{2}

Explanation:

A Length of wire $= 24 a$
Side of hexagon $= a$
Total length of hexagon (circumference) $= 6 a$
Let, number of turns in hexagon $= n$
$24 a = n \times 6 a$
$n = 4$
Area of hexagon $(A) = \frac{3 \sqrt{3}}{2} x^{2} (x =$ side of hexagon $)$
$A = \frac{3 \sqrt{3} a^{2}}{2}$
$∴$ Magnetic moment $(M) = nIA$
$M = 4 \times I \times \frac{3 \sqrt{3} a^{2}}{2}$
$M = 6 \sqrt{3} I a^{2}$

JIPMER-2018

Moving Charges & Magnetism

153926 A moving coil galvanometer has a rectangular wire coil of enclosed area $0.001 m^{2}$ and 500 turns. The coil operates in a radial magnetic field of $0.2 T$ and carries a current of $6 π \times 10^{- 8} A$ . If the torsional spring constant is $6 \times 10^{- 7} N - m / rad$ , then the angular deflection of the coil in radians is

1

\frac{π}{100}

2

\frac{π}{200}

3

\frac{π}{300}

4

\frac{π}{400}

Explanation:

A Given that, $n = 500$ turns, $B = 0.2 T$
Area $(A) = 0.001 m^{2}, I = 6 π \times 10^{- 8} A$
$k = 6 \times 10^{- 7} N - m / rad$
We know that, the expansion of angular deflection of the coil $(θ) = (\frac{nBA}{k}) I$
$θ = \frac{500 \times 0.2 \times 0.001}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = \frac{5 \times 10^{2} \times 2 \times 10^{- 4}}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = π \times 10^{3} \times 10^{7} \times 10^{- 12} = π \times 10^{- 2}$
$θ = \frac{π}{100}$

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153927 The magnetic field in a travelling electromagnetic wave has a peak value of 20nT. The peak value of electric field strength is

1

3 V / m

2

6 V / m

3

9 V / m

4

12 V / m

Explanation:

B Given that,
$B_{0} = 20 nT = 20 \times 10^{- 9}$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 20 \times 10^{- 9}$
$E_{o} = \frac{6 V}{m}$

VITEEE-2018

Moving Charges & Magnetism

153928 A coil in the shape of an equilateral triangle of side $l$ is suspended between two pole pieces of a permanent magnet, such that magnetic field, $B$ 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

2 {(\frac{τ}{\sqrt{3} BI})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{BI})

3

2 {(\frac{τ}{BI})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{BI}

Explanation:

A
Magnetic moment $(\vec{M}) = \vec{I} \times \vec{A}$
Torque $(\vec{τ}) = \vec{M} \times \vec{B}$
$τ = MB \sin θ$
$(θ = 90^{\circ})$
$τ = MB \sin 90^{\circ}$
$τ = MB$
$τ = IAB$
$(∵ M = IA)$
Area of equilateral triangle of side length $= \frac{\sqrt{3}}{4} l^{2}$
$\frac{\sqrt{3}}{4} l^{2} = \frac{τ}{IB}$
$l^{2} = \frac{4 τ}{\sqrt{3} IB}$
$l^{2} = 2 {(\frac{τ}{\sqrt{3} IB})}^{1 / 2}$

BCECE-2017

Moving Charges & Magnetism

153924 Two short bar magnets $A$ and $B$ are arranged coaxially. The distance between their centers is $30$ cm. A compass needle placed on their axis at a distance of $6 cm$ from $B$ shows no deflection. The ratio of the magnetic moments of $A$ and $B$ is

1

16 : 1

2

1 : 16

3

64 : 1

4

1 : 64

Explanation:

C Distance between the magnets (d) $= 30 cm$ Compass needle distance from magnet $(B) = 6 cm$ Now, distance of magnet $A$ from needle, $d_{1} = 30 - 6 = 24 cm$
and distance of magnet $B$ from needle $(d_{2}) = 6 cm$
$∴$ Magnetic field on a axial line is given as,
$B = \frac{μ_{o}}{4 π} \cdot \frac{2 M}{d^{3}}$
Hence, the ratio of the magnetic moment for $A$ and $B$
$\frac{M_{1}}{d_{1}^{3}} = \frac{M_{2}}{d_{2}^{3}}$
$\frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{24}{6})}^{3}$
$\frac{M_{1}}{M_{2}} = 4^{3} = 64$
So, the ratio of the magnetic moment of $A$ and $B$ is $64 : 1$

AP EAMCET (22.04.2019) Shift-II

Moving Charges & Magnetism

153925 A regular hexagon of side a. A wire of length 24a is coiled on that hexagon. If current in hexagon is $I$ , then find the magnetic moment.

1

6 \sqrt{3} {Ia}^{2}

2

3 \sqrt{3} {Ia}^{2}

3

\frac{3 \sqrt{3}}{2} {Is}^{2}

4

6 {Ia}^{2}

Explanation:

A Length of wire $= 24 a$
Side of hexagon $= a$
Total length of hexagon (circumference) $= 6 a$
Let, number of turns in hexagon $= n$
$24 a = n \times 6 a$
$n = 4$
Area of hexagon $(A) = \frac{3 \sqrt{3}}{2} x^{2} (x =$ side of hexagon $)$
$A = \frac{3 \sqrt{3} a^{2}}{2}$
$∴$ Magnetic moment $(M) = nIA$
$M = 4 \times I \times \frac{3 \sqrt{3} a^{2}}{2}$
$M = 6 \sqrt{3} I a^{2}$

JIPMER-2018

Moving Charges & Magnetism

153926 A moving coil galvanometer has a rectangular wire coil of enclosed area $0.001 m^{2}$ and 500 turns. The coil operates in a radial magnetic field of $0.2 T$ and carries a current of $6 π \times 10^{- 8} A$ . If the torsional spring constant is $6 \times 10^{- 7} N - m / rad$ , then the angular deflection of the coil in radians is

1

\frac{π}{100}

2

\frac{π}{200}

3

\frac{π}{300}

4

\frac{π}{400}

Explanation:

A Given that, $n = 500$ turns, $B = 0.2 T$
Area $(A) = 0.001 m^{2}, I = 6 π \times 10^{- 8} A$
$k = 6 \times 10^{- 7} N - m / rad$
We know that, the expansion of angular deflection of the coil $(θ) = (\frac{nBA}{k}) I$
$θ = \frac{500 \times 0.2 \times 0.001}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = \frac{5 \times 10^{2} \times 2 \times 10^{- 4}}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = π \times 10^{3} \times 10^{7} \times 10^{- 12} = π \times 10^{- 2}$
$θ = \frac{π}{100}$

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153927 The magnetic field in a travelling electromagnetic wave has a peak value of 20nT. The peak value of electric field strength is

1

3 V / m

2

6 V / m

3

9 V / m

4

12 V / m

Explanation:

B Given that,
$B_{0} = 20 nT = 20 \times 10^{- 9}$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 20 \times 10^{- 9}$
$E_{o} = \frac{6 V}{m}$

VITEEE-2018

Moving Charges & Magnetism

153928 A coil in the shape of an equilateral triangle of side $l$ is suspended between two pole pieces of a permanent magnet, such that magnetic field, $B$ 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

2 {(\frac{τ}{\sqrt{3} BI})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{BI})

3

2 {(\frac{τ}{BI})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{BI}

Explanation:

A
Magnetic moment $(\vec{M}) = \vec{I} \times \vec{A}$
Torque $(\vec{τ}) = \vec{M} \times \vec{B}$
$τ = MB \sin θ$
$(θ = 90^{\circ})$
$τ = MB \sin 90^{\circ}$
$τ = MB$
$τ = IAB$
$(∵ M = IA)$
Area of equilateral triangle of side length $= \frac{\sqrt{3}}{4} l^{2}$
$\frac{\sqrt{3}}{4} l^{2} = \frac{τ}{IB}$
$l^{2} = \frac{4 τ}{\sqrt{3} IB}$
$l^{2} = 2 {(\frac{τ}{\sqrt{3} IB})}^{1 / 2}$

BCECE-2017

Moving Charges & Magnetism

153924 Two short bar magnets $A$ and $B$ are arranged coaxially. The distance between their centers is $30$ cm. A compass needle placed on their axis at a distance of $6 cm$ from $B$ shows no deflection. The ratio of the magnetic moments of $A$ and $B$ is

1

16 : 1

2

1 : 16

3

64 : 1

4

1 : 64

Explanation:

C Distance between the magnets (d) $= 30 cm$ Compass needle distance from magnet $(B) = 6 cm$ Now, distance of magnet $A$ from needle, $d_{1} = 30 - 6 = 24 cm$
and distance of magnet $B$ from needle $(d_{2}) = 6 cm$
$∴$ Magnetic field on a axial line is given as,
$B = \frac{μ_{o}}{4 π} \cdot \frac{2 M}{d^{3}}$
Hence, the ratio of the magnetic moment for $A$ and $B$
$\frac{M_{1}}{d_{1}^{3}} = \frac{M_{2}}{d_{2}^{3}}$
$\frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{24}{6})}^{3}$
$\frac{M_{1}}{M_{2}} = 4^{3} = 64$
So, the ratio of the magnetic moment of $A$ and $B$ is $64 : 1$

AP EAMCET (22.04.2019) Shift-II

Moving Charges & Magnetism

153925 A regular hexagon of side a. A wire of length 24a is coiled on that hexagon. If current in hexagon is $I$ , then find the magnetic moment.

1

6 \sqrt{3} {Ia}^{2}

2

3 \sqrt{3} {Ia}^{2}

3

\frac{3 \sqrt{3}}{2} {Is}^{2}

4

6 {Ia}^{2}

Explanation:

A Length of wire $= 24 a$
Side of hexagon $= a$
Total length of hexagon (circumference) $= 6 a$
Let, number of turns in hexagon $= n$
$24 a = n \times 6 a$
$n = 4$
Area of hexagon $(A) = \frac{3 \sqrt{3}}{2} x^{2} (x =$ side of hexagon $)$
$A = \frac{3 \sqrt{3} a^{2}}{2}$
$∴$ Magnetic moment $(M) = nIA$
$M = 4 \times I \times \frac{3 \sqrt{3} a^{2}}{2}$
$M = 6 \sqrt{3} I a^{2}$

JIPMER-2018

Moving Charges & Magnetism

153926 A moving coil galvanometer has a rectangular wire coil of enclosed area $0.001 m^{2}$ and 500 turns. The coil operates in a radial magnetic field of $0.2 T$ and carries a current of $6 π \times 10^{- 8} A$ . If the torsional spring constant is $6 \times 10^{- 7} N - m / rad$ , then the angular deflection of the coil in radians is

1

\frac{π}{100}

2

\frac{π}{200}

3

\frac{π}{300}

4

\frac{π}{400}

Explanation:

A Given that, $n = 500$ turns, $B = 0.2 T$
Area $(A) = 0.001 m^{2}, I = 6 π \times 10^{- 8} A$
$k = 6 \times 10^{- 7} N - m / rad$
We know that, the expansion of angular deflection of the coil $(θ) = (\frac{nBA}{k}) I$
$θ = \frac{500 \times 0.2 \times 0.001}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = \frac{5 \times 10^{2} \times 2 \times 10^{- 4}}{6 \times 10^{- 7}} \times 6 π \times 10^{- 8}$
$θ = π \times 10^{3} \times 10^{7} \times 10^{- 12} = π \times 10^{- 2}$
$θ = \frac{π}{100}$

TS- EAMCET-04.05.2018

Moving Charges & Magnetism

153927 The magnetic field in a travelling electromagnetic wave has a peak value of 20nT. The peak value of electric field strength is

1

3 V / m

2

6 V / m

3

9 V / m

4

12 V / m

Explanation:

B Given that,
$B_{0} = 20 nT = 20 \times 10^{- 9}$
Speed of light $(c) = 3 \times 10^{8} m / s$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 20 \times 10^{- 9}$
$E_{o} = \frac{6 V}{m}$

VITEEE-2018

Moving Charges & Magnetism

153928 A coil in the shape of an equilateral triangle of side $l$ is suspended between two pole pieces of a permanent magnet, such that magnetic field, $B$ 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

2 {(\frac{τ}{\sqrt{3} BI})}^{1 / 2}

2

\frac{2}{\sqrt{3}} (\frac{τ}{BI})

3

2 {(\frac{τ}{BI})}^{1 / 2}

4

\frac{1}{\sqrt{3}} \frac{τ}{BI}

Explanation:

A
Magnetic moment $(\vec{M}) = \vec{I} \times \vec{A}$
Torque $(\vec{τ}) = \vec{M} \times \vec{B}$
$τ = MB \sin θ$
$(θ = 90^{\circ})$
$τ = MB \sin 90^{\circ}$
$τ = MB$
$τ = IAB$
$(∵ M = IA)$
Area of equilateral triangle of side length $= \frac{\sqrt{3}}{4} l^{2}$
$\frac{\sqrt{3}}{4} l^{2} = \frac{τ}{IB}$
$l^{2} = \frac{4 τ}{\sqrt{3} IB}$
$l^{2} = 2 {(\frac{τ}{\sqrt{3} IB})}^{1 / 2}$

BCECE-2017

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Question Bank Series

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