QBManager

Magnetism and Matter

154232 The ratio of magnetic moments of two short magnets which give null deflection in $\tan B$ position at $12 cm$ and $18 cm$ from the centre of a deflection magnetometer is

1

2 : 3

2

8 : 27

3

27 : 8

4

4 : 9

Explanation:

B Distance of first magnet $(d_{1}) = 12 cm$ Distance of second magnet $(d_{2}) = 18 cm$
$\Rightarrow B = \frac{μ_{o}}{4 π} \frac{2 M}{d^{3}}$
So, ratio of magnetic moment
$∴ \frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{12}{18})}^{3} = \frac{8}{27}$
$\frac{M_{1}}{M_{2}} = \frac{8}{27}$
$M_{1} : M_{2} = 8 : 27$

Manipal UGET-2012

Magnetism and Matter

154233 The magnetic needle of a tangent galvanometer is deflected at angle of $30^{\circ}$ due to a current in its coil. The, horizontal component of earth's magnetic field is $0.34 \times 10^{- 4} T$ then magnetic filed at the centre of the coil due to current.

1

1.96 \times 10^{- 5} T

2

1.96 \times 10^{- 4} T

3

1.96 \times 10^{4} T

4

1.96 \times 10^{5} T

Explanation:

A Given that,
Angle of deflection $(θ) = 30^{\circ}$
Horizontal magnetic field $(B_{H}) = 0.34 \times 10^{- 4} T$
Since we know that,
Magnetic field $B = B_{H} \tan θ = 0.34 \times 10^{- 4} \times \tan 30^{\circ}$
$= 0.34 \times 10^{- 4} \times \frac{1}{\sqrt{3}} = 0.196 \times 10^{- 4}$
$= 1.96 \times 10^{- 5} T$

CG PET- 2008

Magnetism and Matter

154234 In a tangent galvanometer a current of $0.1 A$ produces a deflection of $30^{\circ}$ . The current required to produce a deflection of $60^{\circ}$ will be

1

0.6 A

2

0.5 A

3

0.4 A

4

0.3 A

Explanation:

D Given that, $i_{1} = 0.1 A, θ_{1} = 30^{\circ}, θ_{2} = 60^{\circ}$
We know that, magnetic field center
$B = B_{H} \tan θ$
$\frac{μ_{o} I}{2 r} = B_{H} \tan θ (∵ B = \frac{μ I}{2 r})$
$I \propto \tan θ$
So, $\frac{I_{2}}{I_{1}} = \frac{\tan θ_{2}}{\tan θ_{1}}$
$\frac{I_{2}}{0.1} = \frac{\tan 60^{\circ}}{\tan 30^{\circ}}$
$I_{2} = \frac{\tan 60^{\circ} \times 0.1}{\tan 30^{\circ}}$
$I_{2} = \frac{\sqrt{3}}{1 / \sqrt{3}} \times 0.1$
$I_{2} = 0.3 A$

SRMJEEE - 2014

Magnetism and Matter

154236 A certain amount of current when flowing in a properly set tangent galvanometer, produces a deflection of $45^{\circ}$ . If the current be reduced by a factor of $\sqrt{3}$ , the deflection would

1 decrease by

30^{\circ}

2 decrease by

15^{\circ}

3 increased by

15^{\circ}

4 increased by

30^{\circ}

Explanation:

B Given that, $I_{2} = \frac{I_{1}}{\sqrt{3}}$
Angle of deflection $(θ_{1}) = 45^{\circ}$
We know that,
$B = B_{H} \tan θ$
$\frac{μ_{0} I}{2 r} = B_{H} \tan θ$
I $\propto \tan θ$
Hence, $\frac{I_{1}}{I_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}}$
$\frac{I_{1}}{I_{1} / \sqrt{3}} = \frac{\tan 45^{\circ}}{\tan θ_{2}}$
$\tan θ_{2} = \frac{\tan 45^{\circ}}{\sqrt{3}}$
$\tan θ_{2} = \frac{1}{\sqrt{3}}$
$θ_{2} = \tan^{- 1} (\frac{1}{\sqrt{3}})$
$θ_{2} = 30^{\circ}$
Hence deflection decrease by,
$θ_{1} - θ_{2} = 45^{\circ} - 30^{\circ} = 15^{\circ} ($ decrease by $15^{\circ})$

CG PET- 2004

Magnetism and Matter

154232 The ratio of magnetic moments of two short magnets which give null deflection in $\tan B$ position at $12 cm$ and $18 cm$ from the centre of a deflection magnetometer is

1

2 : 3

2

8 : 27

3

27 : 8

4

4 : 9

Explanation:

B Distance of first magnet $(d_{1}) = 12 cm$ Distance of second magnet $(d_{2}) = 18 cm$
$\Rightarrow B = \frac{μ_{o}}{4 π} \frac{2 M}{d^{3}}$
So, ratio of magnetic moment
$∴ \frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{12}{18})}^{3} = \frac{8}{27}$
$\frac{M_{1}}{M_{2}} = \frac{8}{27}$
$M_{1} : M_{2} = 8 : 27$

Manipal UGET-2012

Magnetism and Matter

154233 The magnetic needle of a tangent galvanometer is deflected at angle of $30^{\circ}$ due to a current in its coil. The, horizontal component of earth's magnetic field is $0.34 \times 10^{- 4} T$ then magnetic filed at the centre of the coil due to current.

1

1.96 \times 10^{- 5} T

2

1.96 \times 10^{- 4} T

3

1.96 \times 10^{4} T

4

1.96 \times 10^{5} T

Explanation:

A Given that,
Angle of deflection $(θ) = 30^{\circ}$
Horizontal magnetic field $(B_{H}) = 0.34 \times 10^{- 4} T$
Since we know that,
Magnetic field $B = B_{H} \tan θ = 0.34 \times 10^{- 4} \times \tan 30^{\circ}$
$= 0.34 \times 10^{- 4} \times \frac{1}{\sqrt{3}} = 0.196 \times 10^{- 4}$
$= 1.96 \times 10^{- 5} T$

CG PET- 2008

Magnetism and Matter

154234 In a tangent galvanometer a current of $0.1 A$ produces a deflection of $30^{\circ}$ . The current required to produce a deflection of $60^{\circ}$ will be

1

0.6 A

2

0.5 A

3

0.4 A

4

0.3 A

Explanation:

D Given that, $i_{1} = 0.1 A, θ_{1} = 30^{\circ}, θ_{2} = 60^{\circ}$
We know that, magnetic field center
$B = B_{H} \tan θ$
$\frac{μ_{o} I}{2 r} = B_{H} \tan θ (∵ B = \frac{μ I}{2 r})$
$I \propto \tan θ$
So, $\frac{I_{2}}{I_{1}} = \frac{\tan θ_{2}}{\tan θ_{1}}$
$\frac{I_{2}}{0.1} = \frac{\tan 60^{\circ}}{\tan 30^{\circ}}$
$I_{2} = \frac{\tan 60^{\circ} \times 0.1}{\tan 30^{\circ}}$
$I_{2} = \frac{\sqrt{3}}{1 / \sqrt{3}} \times 0.1$
$I_{2} = 0.3 A$

SRMJEEE - 2014

Magnetism and Matter

154236 A certain amount of current when flowing in a properly set tangent galvanometer, produces a deflection of $45^{\circ}$ . If the current be reduced by a factor of $\sqrt{3}$ , the deflection would

1 decrease by

30^{\circ}

2 decrease by

15^{\circ}

3 increased by

15^{\circ}

4 increased by

30^{\circ}

Explanation:

B Given that, $I_{2} = \frac{I_{1}}{\sqrt{3}}$
Angle of deflection $(θ_{1}) = 45^{\circ}$
We know that,
$B = B_{H} \tan θ$
$\frac{μ_{0} I}{2 r} = B_{H} \tan θ$
I $\propto \tan θ$
Hence, $\frac{I_{1}}{I_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}}$
$\frac{I_{1}}{I_{1} / \sqrt{3}} = \frac{\tan 45^{\circ}}{\tan θ_{2}}$
$\tan θ_{2} = \frac{\tan 45^{\circ}}{\sqrt{3}}$
$\tan θ_{2} = \frac{1}{\sqrt{3}}$
$θ_{2} = \tan^{- 1} (\frac{1}{\sqrt{3}})$
$θ_{2} = 30^{\circ}$
Hence deflection decrease by,
$θ_{1} - θ_{2} = 45^{\circ} - 30^{\circ} = 15^{\circ} ($ decrease by $15^{\circ})$

CG PET- 2004

Magnetism and Matter

154232 The ratio of magnetic moments of two short magnets which give null deflection in $\tan B$ position at $12 cm$ and $18 cm$ from the centre of a deflection magnetometer is

1

2 : 3

2

8 : 27

3

27 : 8

4

4 : 9

Explanation:

B Distance of first magnet $(d_{1}) = 12 cm$ Distance of second magnet $(d_{2}) = 18 cm$
$\Rightarrow B = \frac{μ_{o}}{4 π} \frac{2 M}{d^{3}}$
So, ratio of magnetic moment
$∴ \frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{12}{18})}^{3} = \frac{8}{27}$
$\frac{M_{1}}{M_{2}} = \frac{8}{27}$
$M_{1} : M_{2} = 8 : 27$

Manipal UGET-2012

Magnetism and Matter

154233 The magnetic needle of a tangent galvanometer is deflected at angle of $30^{\circ}$ due to a current in its coil. The, horizontal component of earth's magnetic field is $0.34 \times 10^{- 4} T$ then magnetic filed at the centre of the coil due to current.

1

1.96 \times 10^{- 5} T

2

1.96 \times 10^{- 4} T

3

1.96 \times 10^{4} T

4

1.96 \times 10^{5} T

Explanation:

A Given that,
Angle of deflection $(θ) = 30^{\circ}$
Horizontal magnetic field $(B_{H}) = 0.34 \times 10^{- 4} T$
Since we know that,
Magnetic field $B = B_{H} \tan θ = 0.34 \times 10^{- 4} \times \tan 30^{\circ}$
$= 0.34 \times 10^{- 4} \times \frac{1}{\sqrt{3}} = 0.196 \times 10^{- 4}$
$= 1.96 \times 10^{- 5} T$

CG PET- 2008

Magnetism and Matter

154234 In a tangent galvanometer a current of $0.1 A$ produces a deflection of $30^{\circ}$ . The current required to produce a deflection of $60^{\circ}$ will be

1

0.6 A

2

0.5 A

3

0.4 A

4

0.3 A

Explanation:

D Given that, $i_{1} = 0.1 A, θ_{1} = 30^{\circ}, θ_{2} = 60^{\circ}$
We know that, magnetic field center
$B = B_{H} \tan θ$
$\frac{μ_{o} I}{2 r} = B_{H} \tan θ (∵ B = \frac{μ I}{2 r})$
$I \propto \tan θ$
So, $\frac{I_{2}}{I_{1}} = \frac{\tan θ_{2}}{\tan θ_{1}}$
$\frac{I_{2}}{0.1} = \frac{\tan 60^{\circ}}{\tan 30^{\circ}}$
$I_{2} = \frac{\tan 60^{\circ} \times 0.1}{\tan 30^{\circ}}$
$I_{2} = \frac{\sqrt{3}}{1 / \sqrt{3}} \times 0.1$
$I_{2} = 0.3 A$

SRMJEEE - 2014

Magnetism and Matter

154236 A certain amount of current when flowing in a properly set tangent galvanometer, produces a deflection of $45^{\circ}$ . If the current be reduced by a factor of $\sqrt{3}$ , the deflection would

1 decrease by

30^{\circ}

2 decrease by

15^{\circ}

3 increased by

15^{\circ}

4 increased by

30^{\circ}

Explanation:

B Given that, $I_{2} = \frac{I_{1}}{\sqrt{3}}$
Angle of deflection $(θ_{1}) = 45^{\circ}$
We know that,
$B = B_{H} \tan θ$
$\frac{μ_{0} I}{2 r} = B_{H} \tan θ$
I $\propto \tan θ$
Hence, $\frac{I_{1}}{I_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}}$
$\frac{I_{1}}{I_{1} / \sqrt{3}} = \frac{\tan 45^{\circ}}{\tan θ_{2}}$
$\tan θ_{2} = \frac{\tan 45^{\circ}}{\sqrt{3}}$
$\tan θ_{2} = \frac{1}{\sqrt{3}}$
$θ_{2} = \tan^{- 1} (\frac{1}{\sqrt{3}})$
$θ_{2} = 30^{\circ}$
Hence deflection decrease by,
$θ_{1} - θ_{2} = 45^{\circ} - 30^{\circ} = 15^{\circ} ($ decrease by $15^{\circ})$

CG PET- 2004

Magnetism and Matter

154232 The ratio of magnetic moments of two short magnets which give null deflection in $\tan B$ position at $12 cm$ and $18 cm$ from the centre of a deflection magnetometer is

1

2 : 3

2

8 : 27

3

27 : 8

4

4 : 9

Explanation:

B Distance of first magnet $(d_{1}) = 12 cm$ Distance of second magnet $(d_{2}) = 18 cm$
$\Rightarrow B = \frac{μ_{o}}{4 π} \frac{2 M}{d^{3}}$
So, ratio of magnetic moment
$∴ \frac{M_{1}}{M_{2}} = {(\frac{d_{1}}{d_{2}})}^{3}$
$\frac{M_{1}}{M_{2}} = {(\frac{12}{18})}^{3} = \frac{8}{27}$
$\frac{M_{1}}{M_{2}} = \frac{8}{27}$
$M_{1} : M_{2} = 8 : 27$

Manipal UGET-2012

Magnetism and Matter

154233 The magnetic needle of a tangent galvanometer is deflected at angle of $30^{\circ}$ due to a current in its coil. The, horizontal component of earth's magnetic field is $0.34 \times 10^{- 4} T$ then magnetic filed at the centre of the coil due to current.

1

1.96 \times 10^{- 5} T

2

1.96 \times 10^{- 4} T

3

1.96 \times 10^{4} T

4

1.96 \times 10^{5} T

Explanation:

A Given that,
Angle of deflection $(θ) = 30^{\circ}$
Horizontal magnetic field $(B_{H}) = 0.34 \times 10^{- 4} T$
Since we know that,
Magnetic field $B = B_{H} \tan θ = 0.34 \times 10^{- 4} \times \tan 30^{\circ}$
$= 0.34 \times 10^{- 4} \times \frac{1}{\sqrt{3}} = 0.196 \times 10^{- 4}$
$= 1.96 \times 10^{- 5} T$

CG PET- 2008

Magnetism and Matter

154234 In a tangent galvanometer a current of $0.1 A$ produces a deflection of $30^{\circ}$ . The current required to produce a deflection of $60^{\circ}$ will be

1

0.6 A

2

0.5 A

3

0.4 A

4

0.3 A

Explanation:

D Given that, $i_{1} = 0.1 A, θ_{1} = 30^{\circ}, θ_{2} = 60^{\circ}$
We know that, magnetic field center
$B = B_{H} \tan θ$
$\frac{μ_{o} I}{2 r} = B_{H} \tan θ (∵ B = \frac{μ I}{2 r})$
$I \propto \tan θ$
So, $\frac{I_{2}}{I_{1}} = \frac{\tan θ_{2}}{\tan θ_{1}}$
$\frac{I_{2}}{0.1} = \frac{\tan 60^{\circ}}{\tan 30^{\circ}}$
$I_{2} = \frac{\tan 60^{\circ} \times 0.1}{\tan 30^{\circ}}$
$I_{2} = \frac{\sqrt{3}}{1 / \sqrt{3}} \times 0.1$
$I_{2} = 0.3 A$

SRMJEEE - 2014

Magnetism and Matter

154236 A certain amount of current when flowing in a properly set tangent galvanometer, produces a deflection of $45^{\circ}$ . If the current be reduced by a factor of $\sqrt{3}$ , the deflection would

1 decrease by

30^{\circ}

2 decrease by

15^{\circ}

3 increased by

15^{\circ}

4 increased by

30^{\circ}

Explanation:

B Given that, $I_{2} = \frac{I_{1}}{\sqrt{3}}$
Angle of deflection $(θ_{1}) = 45^{\circ}$
We know that,
$B = B_{H} \tan θ$
$\frac{μ_{0} I}{2 r} = B_{H} \tan θ$
I $\propto \tan θ$
Hence, $\frac{I_{1}}{I_{2}} = \frac{\tan θ_{1}}{\tan θ_{2}}$
$\frac{I_{1}}{I_{1} / \sqrt{3}} = \frac{\tan 45^{\circ}}{\tan θ_{2}}$
$\tan θ_{2} = \frac{\tan 45^{\circ}}{\sqrt{3}}$
$\tan θ_{2} = \frac{1}{\sqrt{3}}$
$θ_{2} = \tan^{- 1} (\frac{1}{\sqrt{3}})$
$θ_{2} = 30^{\circ}$
Hence deflection decrease by,
$θ_{1} - θ_{2} = 45^{\circ} - 30^{\circ} = 15^{\circ} ($ decrease by $15^{\circ})$

CG PET- 2004