Answer
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Hint:Use the formula for the horizontal component of Earth’s magnetic field. This formula gives the relation between the horizontal component of Earth’s magnetic field, magnetic field of the Earth and the angle between the magnetic field of the Earth and horizontal component of Earth’s magnetic field. Substitute the values given in the question and determine the value of the magnetic field at which the magnetic needle is placed.
Formula used:
The horizontal component of the Earth’s magnetic field \[H\] is given by
\[H = {B_e}\cos \theta \] …… (1)
Here, \[{B_e}\] is the magnitude of the Earth’s magnetic field and \[\theta \] is the angle made by the horizontal component of the Earth’s magnetic field with the magnetic field of the Earth.
Complete step by step answer:
We have given that the magnetic needle is rotated freely in a vertical plane parallel to the magnetic meridian with its north tip pointing down at \[22^\circ \] with the horizontal.
\[\theta = 22^\circ \]
We have given that the horizontal component of the Earth’s magnetic field at the place where the magnetic needle is free to rotate is \[0.35G\].
\[H = 0.35G\]
We have asked to determine the magnitude of the Earth’s magnetic field at the place where the needle is free to rotate.
Rearrange the equation (1) for the magnitude of the magnetic field \[{B_e}\].
\[{B_e} = \dfrac{H}{{\cos \theta }}\]
Substitute \[0.35G\] for \[H\] and \[22^\circ \] for \[\theta \] in the above equation.
\[{B_e} = \dfrac{{0.35G}}{{\cos 22^\circ }}\]
\[ \Rightarrow {B_e} = \dfrac{{0.35G}}{{0.9271}}\]
\[ \Rightarrow {B_e} = 0.37G\]
\[ \therefore {B_e} \approx 0.36G\]
Therefore, the magnetic field of the Earth at the place where the needle is free to rotate is \[0.36G\].
Hence, the correct option is B.
Note: The students should keep in mind that the actual value of the magnetic field at a place where the magnetic needle is free to rotate is 0.377G. But this value is not given in the option. So, we have rounded the value of the magnetic field at the place where the magnetic needle is free to rotate to the nearest value 0.36G which is given in the option.
Formula used:
The horizontal component of the Earth’s magnetic field \[H\] is given by
\[H = {B_e}\cos \theta \] …… (1)
Here, \[{B_e}\] is the magnitude of the Earth’s magnetic field and \[\theta \] is the angle made by the horizontal component of the Earth’s magnetic field with the magnetic field of the Earth.
Complete step by step answer:
We have given that the magnetic needle is rotated freely in a vertical plane parallel to the magnetic meridian with its north tip pointing down at \[22^\circ \] with the horizontal.
\[\theta = 22^\circ \]
We have given that the horizontal component of the Earth’s magnetic field at the place where the magnetic needle is free to rotate is \[0.35G\].
\[H = 0.35G\]
We have asked to determine the magnitude of the Earth’s magnetic field at the place where the needle is free to rotate.
Rearrange the equation (1) for the magnitude of the magnetic field \[{B_e}\].
\[{B_e} = \dfrac{H}{{\cos \theta }}\]
Substitute \[0.35G\] for \[H\] and \[22^\circ \] for \[\theta \] in the above equation.
\[{B_e} = \dfrac{{0.35G}}{{\cos 22^\circ }}\]
\[ \Rightarrow {B_e} = \dfrac{{0.35G}}{{0.9271}}\]
\[ \Rightarrow {B_e} = 0.37G\]
\[ \therefore {B_e} \approx 0.36G\]
Therefore, the magnetic field of the Earth at the place where the needle is free to rotate is \[0.36G\].
Hence, the correct option is B.
Note: The students should keep in mind that the actual value of the magnetic field at a place where the magnetic needle is free to rotate is 0.377G. But this value is not given in the option. So, we have rounded the value of the magnetic field at the place where the magnetic needle is free to rotate to the nearest value 0.36G which is given in the option.