154116
The angle of dip at a place where horizontal and vertical components of Earth's magnetic field are equal is
1 $45^{\circ}$
2 $30^{\circ}$
3 $0^{\circ}$
4 $60^{\circ}$
5 $90^{\circ}$
Explanation:
A We know that, Vertical component of earth's magnetic field is , $\mathrm{B}_{\mathrm{V}}=\mathrm{B} \sin \theta$ Horizontal component of earth's magnetic field is, $\mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \theta$ According to question, $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{V}}$ $\mathrm{B} \cos \theta=\mathrm{B} \sin \theta$ $\frac{\mathrm{B} \sin \theta}{\mathrm{B} \cos \theta}=1$ $\tan \theta=1$ $\tan \theta=\tan 45^{\circ}$ $\theta=45^{\circ}$
Kerala CEE- 2021
Magnetism and Matter
154117
The angle of dip at a place is $30^{\circ}$. If horizontal component of earth's magnetic field is $H$, the total field intensity is
1 $\frac{\mathrm{H}}{\sqrt{2}}$
2 $\frac{2 \mathrm{H}}{\sqrt{3}}$
3 $\mathrm{H} \sqrt{2}$
4 $\mathrm{H} \sqrt{3}$
Explanation:
B Given, Angle of dip $(\theta)=30^{\circ}$ Horizontal component of earth's magnetic field, $\mathrm{B}_{\mathrm{H}}=\mathrm{H}$ We know that, $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{e}} \cos \theta$ $\mathrm{B}_{\mathrm{e}}=\frac{\mathrm{B}_{\mathrm{H}}}{\operatorname{Cos} 30^{\circ}}$ $\mathrm{B}_{\mathrm{e}}=\frac{2 \mathrm{H}}{\sqrt{3}}$
UPSEE 2020
Magnetism and Matter
154120
In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $\frac{\sqrt{3}}{5} G$ and the dip angle of a magnetic needle is $30^{\circ}$. What is the magnetic field of the earth at this location?
1 $0.4 \mathrm{G}$
2 $0.2 \mathrm{G}$
3 $\frac{2 \sqrt{3}}{5} \mathrm{G}$
4 $\frac{\sqrt{2}}{5} \mathrm{G}$
Explanation:
A Given that, $\mathrm{B}_{\mathrm{H}}=\frac{\sqrt{3}}{5} \mathrm{G}, \text { dip angle }(\delta)=30^{\circ}$ We know that, $\mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \delta$ $\cos \delta=\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{B}}$ Then, $\cos 30^{\circ}=\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{B}}$ $\frac{\sqrt{3}}{2}=\frac{\frac{\sqrt{3}}{5} \mathrm{G}}{\mathrm{B}}$ $\mathrm{B}=\frac{\sqrt{3}}{5} \times \frac{2}{\sqrt{3}}=\frac{2}{5}$ So, $\mathrm{B}=0.4 \mathrm{G}$
TS EAMCET 29.09.2020
Magnetism and Matter
154121
A coil of 50 turns and $10 \mathrm{~cm}$ diameter has a resistance of $10 \Omega$. What must the potential difference across the coil so as to nullify the earths magnetic field $B_{H}=0.314 \mathrm{G}$ at the centre of the coil?
1 $0.5 \mathrm{~V}$
2 $1.0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.0 \mathrm{~V}$
Explanation:
A Given that, $\mathrm{B}_{\mathrm{H}}=0.314 \mathrm{G}$ $\mathrm{B}_{\mathrm{H}}=0.314 \times 10^{-4} \mathrm{~T}$ The current in coil be 'I' then the magnetic field at the centre of the coil is - $B=\frac{\mu_{0} \mathrm{In}}{2 \mathrm{r}}$ $B=\frac{4 \pi \times 10^{-7} \times \mathrm{I} \times 50}{2 \times 5 \times 10^{-2}}$ $\mathrm{~B}=4 \pi \times 10^{-5} \times 5 \times \mathrm{I}$ $\mathrm{I}=\frac{\mathrm{B}}{4 \pi \times 10^{-5} \times 5}$ $\mathrm{I}=\frac{0.314 \times 10^{-4}}{4 \pi \times 10^{-5} \times 5}$ $\mathrm{I}=\frac{1}{20} \Rightarrow 0.05 \mathrm{~A}$ $\therefore$ Potential difference $(\mathrm{V})=\mathrm{IR}$ $=0.05 \times 10=0.5 \mathrm{~V}$
154116
The angle of dip at a place where horizontal and vertical components of Earth's magnetic field are equal is
1 $45^{\circ}$
2 $30^{\circ}$
3 $0^{\circ}$
4 $60^{\circ}$
5 $90^{\circ}$
Explanation:
A We know that, Vertical component of earth's magnetic field is , $\mathrm{B}_{\mathrm{V}}=\mathrm{B} \sin \theta$ Horizontal component of earth's magnetic field is, $\mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \theta$ According to question, $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{V}}$ $\mathrm{B} \cos \theta=\mathrm{B} \sin \theta$ $\frac{\mathrm{B} \sin \theta}{\mathrm{B} \cos \theta}=1$ $\tan \theta=1$ $\tan \theta=\tan 45^{\circ}$ $\theta=45^{\circ}$
Kerala CEE- 2021
Magnetism and Matter
154117
The angle of dip at a place is $30^{\circ}$. If horizontal component of earth's magnetic field is $H$, the total field intensity is
1 $\frac{\mathrm{H}}{\sqrt{2}}$
2 $\frac{2 \mathrm{H}}{\sqrt{3}}$
3 $\mathrm{H} \sqrt{2}$
4 $\mathrm{H} \sqrt{3}$
Explanation:
B Given, Angle of dip $(\theta)=30^{\circ}$ Horizontal component of earth's magnetic field, $\mathrm{B}_{\mathrm{H}}=\mathrm{H}$ We know that, $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{e}} \cos \theta$ $\mathrm{B}_{\mathrm{e}}=\frac{\mathrm{B}_{\mathrm{H}}}{\operatorname{Cos} 30^{\circ}}$ $\mathrm{B}_{\mathrm{e}}=\frac{2 \mathrm{H}}{\sqrt{3}}$
UPSEE 2020
Magnetism and Matter
154120
In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $\frac{\sqrt{3}}{5} G$ and the dip angle of a magnetic needle is $30^{\circ}$. What is the magnetic field of the earth at this location?
1 $0.4 \mathrm{G}$
2 $0.2 \mathrm{G}$
3 $\frac{2 \sqrt{3}}{5} \mathrm{G}$
4 $\frac{\sqrt{2}}{5} \mathrm{G}$
Explanation:
A Given that, $\mathrm{B}_{\mathrm{H}}=\frac{\sqrt{3}}{5} \mathrm{G}, \text { dip angle }(\delta)=30^{\circ}$ We know that, $\mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \delta$ $\cos \delta=\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{B}}$ Then, $\cos 30^{\circ}=\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{B}}$ $\frac{\sqrt{3}}{2}=\frac{\frac{\sqrt{3}}{5} \mathrm{G}}{\mathrm{B}}$ $\mathrm{B}=\frac{\sqrt{3}}{5} \times \frac{2}{\sqrt{3}}=\frac{2}{5}$ So, $\mathrm{B}=0.4 \mathrm{G}$
TS EAMCET 29.09.2020
Magnetism and Matter
154121
A coil of 50 turns and $10 \mathrm{~cm}$ diameter has a resistance of $10 \Omega$. What must the potential difference across the coil so as to nullify the earths magnetic field $B_{H}=0.314 \mathrm{G}$ at the centre of the coil?
1 $0.5 \mathrm{~V}$
2 $1.0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.0 \mathrm{~V}$
Explanation:
A Given that, $\mathrm{B}_{\mathrm{H}}=0.314 \mathrm{G}$ $\mathrm{B}_{\mathrm{H}}=0.314 \times 10^{-4} \mathrm{~T}$ The current in coil be 'I' then the magnetic field at the centre of the coil is - $B=\frac{\mu_{0} \mathrm{In}}{2 \mathrm{r}}$ $B=\frac{4 \pi \times 10^{-7} \times \mathrm{I} \times 50}{2 \times 5 \times 10^{-2}}$ $\mathrm{~B}=4 \pi \times 10^{-5} \times 5 \times \mathrm{I}$ $\mathrm{I}=\frac{\mathrm{B}}{4 \pi \times 10^{-5} \times 5}$ $\mathrm{I}=\frac{0.314 \times 10^{-4}}{4 \pi \times 10^{-5} \times 5}$ $\mathrm{I}=\frac{1}{20} \Rightarrow 0.05 \mathrm{~A}$ $\therefore$ Potential difference $(\mathrm{V})=\mathrm{IR}$ $=0.05 \times 10=0.5 \mathrm{~V}$
154116
The angle of dip at a place where horizontal and vertical components of Earth's magnetic field are equal is
1 $45^{\circ}$
2 $30^{\circ}$
3 $0^{\circ}$
4 $60^{\circ}$
5 $90^{\circ}$
Explanation:
A We know that, Vertical component of earth's magnetic field is , $\mathrm{B}_{\mathrm{V}}=\mathrm{B} \sin \theta$ Horizontal component of earth's magnetic field is, $\mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \theta$ According to question, $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{V}}$ $\mathrm{B} \cos \theta=\mathrm{B} \sin \theta$ $\frac{\mathrm{B} \sin \theta}{\mathrm{B} \cos \theta}=1$ $\tan \theta=1$ $\tan \theta=\tan 45^{\circ}$ $\theta=45^{\circ}$
Kerala CEE- 2021
Magnetism and Matter
154117
The angle of dip at a place is $30^{\circ}$. If horizontal component of earth's magnetic field is $H$, the total field intensity is
1 $\frac{\mathrm{H}}{\sqrt{2}}$
2 $\frac{2 \mathrm{H}}{\sqrt{3}}$
3 $\mathrm{H} \sqrt{2}$
4 $\mathrm{H} \sqrt{3}$
Explanation:
B Given, Angle of dip $(\theta)=30^{\circ}$ Horizontal component of earth's magnetic field, $\mathrm{B}_{\mathrm{H}}=\mathrm{H}$ We know that, $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{e}} \cos \theta$ $\mathrm{B}_{\mathrm{e}}=\frac{\mathrm{B}_{\mathrm{H}}}{\operatorname{Cos} 30^{\circ}}$ $\mathrm{B}_{\mathrm{e}}=\frac{2 \mathrm{H}}{\sqrt{3}}$
UPSEE 2020
Magnetism and Matter
154120
In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $\frac{\sqrt{3}}{5} G$ and the dip angle of a magnetic needle is $30^{\circ}$. What is the magnetic field of the earth at this location?
1 $0.4 \mathrm{G}$
2 $0.2 \mathrm{G}$
3 $\frac{2 \sqrt{3}}{5} \mathrm{G}$
4 $\frac{\sqrt{2}}{5} \mathrm{G}$
Explanation:
A Given that, $\mathrm{B}_{\mathrm{H}}=\frac{\sqrt{3}}{5} \mathrm{G}, \text { dip angle }(\delta)=30^{\circ}$ We know that, $\mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \delta$ $\cos \delta=\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{B}}$ Then, $\cos 30^{\circ}=\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{B}}$ $\frac{\sqrt{3}}{2}=\frac{\frac{\sqrt{3}}{5} \mathrm{G}}{\mathrm{B}}$ $\mathrm{B}=\frac{\sqrt{3}}{5} \times \frac{2}{\sqrt{3}}=\frac{2}{5}$ So, $\mathrm{B}=0.4 \mathrm{G}$
TS EAMCET 29.09.2020
Magnetism and Matter
154121
A coil of 50 turns and $10 \mathrm{~cm}$ diameter has a resistance of $10 \Omega$. What must the potential difference across the coil so as to nullify the earths magnetic field $B_{H}=0.314 \mathrm{G}$ at the centre of the coil?
1 $0.5 \mathrm{~V}$
2 $1.0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.0 \mathrm{~V}$
Explanation:
A Given that, $\mathrm{B}_{\mathrm{H}}=0.314 \mathrm{G}$ $\mathrm{B}_{\mathrm{H}}=0.314 \times 10^{-4} \mathrm{~T}$ The current in coil be 'I' then the magnetic field at the centre of the coil is - $B=\frac{\mu_{0} \mathrm{In}}{2 \mathrm{r}}$ $B=\frac{4 \pi \times 10^{-7} \times \mathrm{I} \times 50}{2 \times 5 \times 10^{-2}}$ $\mathrm{~B}=4 \pi \times 10^{-5} \times 5 \times \mathrm{I}$ $\mathrm{I}=\frac{\mathrm{B}}{4 \pi \times 10^{-5} \times 5}$ $\mathrm{I}=\frac{0.314 \times 10^{-4}}{4 \pi \times 10^{-5} \times 5}$ $\mathrm{I}=\frac{1}{20} \Rightarrow 0.05 \mathrm{~A}$ $\therefore$ Potential difference $(\mathrm{V})=\mathrm{IR}$ $=0.05 \times 10=0.5 \mathrm{~V}$
154116
The angle of dip at a place where horizontal and vertical components of Earth's magnetic field are equal is
1 $45^{\circ}$
2 $30^{\circ}$
3 $0^{\circ}$
4 $60^{\circ}$
5 $90^{\circ}$
Explanation:
A We know that, Vertical component of earth's magnetic field is , $\mathrm{B}_{\mathrm{V}}=\mathrm{B} \sin \theta$ Horizontal component of earth's magnetic field is, $\mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \theta$ According to question, $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{V}}$ $\mathrm{B} \cos \theta=\mathrm{B} \sin \theta$ $\frac{\mathrm{B} \sin \theta}{\mathrm{B} \cos \theta}=1$ $\tan \theta=1$ $\tan \theta=\tan 45^{\circ}$ $\theta=45^{\circ}$
Kerala CEE- 2021
Magnetism and Matter
154117
The angle of dip at a place is $30^{\circ}$. If horizontal component of earth's magnetic field is $H$, the total field intensity is
1 $\frac{\mathrm{H}}{\sqrt{2}}$
2 $\frac{2 \mathrm{H}}{\sqrt{3}}$
3 $\mathrm{H} \sqrt{2}$
4 $\mathrm{H} \sqrt{3}$
Explanation:
B Given, Angle of dip $(\theta)=30^{\circ}$ Horizontal component of earth's magnetic field, $\mathrm{B}_{\mathrm{H}}=\mathrm{H}$ We know that, $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{e}} \cos \theta$ $\mathrm{B}_{\mathrm{e}}=\frac{\mathrm{B}_{\mathrm{H}}}{\operatorname{Cos} 30^{\circ}}$ $\mathrm{B}_{\mathrm{e}}=\frac{2 \mathrm{H}}{\sqrt{3}}$
UPSEE 2020
Magnetism and Matter
154120
In the magnetic meridian of a certain place, the horizontal component of the earth's magnetic field is $\frac{\sqrt{3}}{5} G$ and the dip angle of a magnetic needle is $30^{\circ}$. What is the magnetic field of the earth at this location?
1 $0.4 \mathrm{G}$
2 $0.2 \mathrm{G}$
3 $\frac{2 \sqrt{3}}{5} \mathrm{G}$
4 $\frac{\sqrt{2}}{5} \mathrm{G}$
Explanation:
A Given that, $\mathrm{B}_{\mathrm{H}}=\frac{\sqrt{3}}{5} \mathrm{G}, \text { dip angle }(\delta)=30^{\circ}$ We know that, $\mathrm{B}_{\mathrm{H}}=\mathrm{B} \cos \delta$ $\cos \delta=\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{B}}$ Then, $\cos 30^{\circ}=\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{B}}$ $\frac{\sqrt{3}}{2}=\frac{\frac{\sqrt{3}}{5} \mathrm{G}}{\mathrm{B}}$ $\mathrm{B}=\frac{\sqrt{3}}{5} \times \frac{2}{\sqrt{3}}=\frac{2}{5}$ So, $\mathrm{B}=0.4 \mathrm{G}$
TS EAMCET 29.09.2020
Magnetism and Matter
154121
A coil of 50 turns and $10 \mathrm{~cm}$ diameter has a resistance of $10 \Omega$. What must the potential difference across the coil so as to nullify the earths magnetic field $B_{H}=0.314 \mathrm{G}$ at the centre of the coil?
1 $0.5 \mathrm{~V}$
2 $1.0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.0 \mathrm{~V}$
Explanation:
A Given that, $\mathrm{B}_{\mathrm{H}}=0.314 \mathrm{G}$ $\mathrm{B}_{\mathrm{H}}=0.314 \times 10^{-4} \mathrm{~T}$ The current in coil be 'I' then the magnetic field at the centre of the coil is - $B=\frac{\mu_{0} \mathrm{In}}{2 \mathrm{r}}$ $B=\frac{4 \pi \times 10^{-7} \times \mathrm{I} \times 50}{2 \times 5 \times 10^{-2}}$ $\mathrm{~B}=4 \pi \times 10^{-5} \times 5 \times \mathrm{I}$ $\mathrm{I}=\frac{\mathrm{B}}{4 \pi \times 10^{-5} \times 5}$ $\mathrm{I}=\frac{0.314 \times 10^{-4}}{4 \pi \times 10^{-5} \times 5}$ $\mathrm{I}=\frac{1}{20} \Rightarrow 0.05 \mathrm{~A}$ $\therefore$ Potential difference $(\mathrm{V})=\mathrm{IR}$ $=0.05 \times 10=0.5 \mathrm{~V}$
