By using R = 2 \(\times\) 10⁴\(\Omega\) in series

F = 0.104 N

Here, n = 2, I = 5.0 A,
B = 0.4 x 10^-3 T, r = ?
Magnetic field at the centre of current loop is
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi nI }{ r } \ or \ r=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi nI }{ B } \) 
\(\therefore \ r={ 10 }^{ -7 }\times 2\times \frac { 22 }{ 7 } \times \frac { 2\times 5.0 }{ 0.4\times { 10 }^{ -3 } } \) 
 = 157.1 x 10^-4m = 1.57 cm  

\({ B }_{ v }=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2M\cos { \theta } }{ { r }^{ 3 } } \)
\({ \ B }_{ H }=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { M\sin { \theta } }{ { r }^{ 3 } } \)
\(B=\sqrt { \begin{matrix} { { B }_{ v } }^{ 2 } & + & { { B }_{ H } }^{ 2 } \end{matrix} } =\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } \left[ 4\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right] \)
\(B=\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } \left[ 3\cos ^{ 2 }{ \theta } +1 \right] \)
Now B will be minimum if \(\cos { \theta } =0 \ or \ \theta ={ 90 }^{ \circ }\)
i.e. B will be minimum at magnetic equator.
(ii) Angle of dip is given by
\(\tan { \delta =\frac { { B }_{ V } }{ { B }_{ H } } } =\frac { 2\cos { \theta } }{ \sin { \theta } } =2\cot { \theta } \)
\(\delta =0,if \ \cot { \theta =0 \ or \ \theta =\frac { \pi }{ 2 } } \)
i.e. angle of dip is zero at magnetic equator(iii) \(\delta ={ 45 }^{ \circ },\cot { \theta =\frac { 1 }{ 2 } } or \ \tan { \theta =2 } \)
\( \theta =\tan ^{ -1 }{ 2is \ the \ locus. } \)

The magnetic Field due to AB and EF is as the direction of length vector is along the radius vector. Also the magnetic field due to BCE and BDE are equal opposite and equal so they cancel the effect of each other. So the net magnetic field at O is 0.

We know the resistance connected to galvanometer to convert it into ammeter is S =(Ig/( I – Ig))xG 
So for higher resistance, the range of I should be small, therefore milliammeter has the higher resistance.

As the magnetic field produced by solenoid is always along its axis, so direction of velocity of proton is along the direction of field, therefore
F = qvB Sin 0 = 0

We know the resistance connected to galvanometer to Convert it into voltmeter is R  - (V/Ig) - G
So if R is higher, range of V will also be higher, so a Voltmeter has the higher resistance.

Spring Will Contract due to the magnetic field produced by the turns of the coil and the weights will be lifted up

Magnetic field induction at a point P in space at a perpendicular distance a from a linear conductor carrying current i is 
\(B=\frac { { \mu }_{ o } }{ 4\pi a } \left( sin{ \phi }_{ 1 }+sin{ \phi }_{ 2 } \right) \) 
Where  \({ \phi }_{ 1 }\) and  \({ \phi }_{ 2 }\) are the angles subtended by the lines joining the ends of linear conductor with point P and perpendicular drawn on the conductor from point P.
In case of infinitely long conductor, 
\( { \phi }_{ 1 }=\frac { \pi }{ 2 } ={ \phi }_{ 2 }\)
\(\\ \therefore \quad \quad B=\frac { { \mu }_{ o } }{ 4\pi } \frac { i }{ a } \left( sin\frac { \pi }{ 2 } +sin\frac { \pi }{ 2 } \right) =\frac { { \mu }_{ o } }{ 4\pi } \frac { 2i }{ a } \)

When a point P lies on the axis of wire conductor in air, then  \(I\overrightarrow { dl } \ and \ \overrightarrow { r } \)  for each element of the straight wire conductor are parallel. Therefore, \(I\overrightarrow { dl } \times \overrightarrow { r } =0,\) so the magnetic field induction at given point is zero.

A current carrying wire produces magnetic field but the wire, which does not carry current has no magnetic field.

Magnetic flux density at a point in a magnetic field means magnetic field induction at that point. It is defined as the force experienced by a unit charge while moving with a unit velocity, perpendicular to the direction of magnetic field at that point. Force experienced by the charged particle having charge q moving with velocity \(\overset { \rightarrow }{ v } \) through a magnetic field \(\overset { \rightarrow }{ B } \) is given by
\( \left| \overset { \rightarrow }{ F } \right| =q\left| \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right| =qvBsin\theta\)  
\(or \ B=\frac { F }{ qvsin\theta } \)
The SI unit of B is tesla, where 1 tesla is the magnetic flux density at a point if 1 coulomb charge while moving with a velocity of 1 ms^-1, perpendicular to a magnetic field experiences a force of 1 N at that point.
The dimensional formula of B
 \(=\frac { \left[ { MLT }^{ -2 } \right] }{ \left[ AT \right] \left[ { LT }^{ -1 } \right] } =\left[ { ML }^{ o }{ T }^{ -2 }{ A }^{ -1 } \right] \)

\(\overset { \rightarrow }{ F } =q\left( \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right) \)  
\(\\ or \ \left| \overset { \rightarrow }{ F } \right| =q\left| \overset { \rightarrow }{ v } \times \overset { \rightarrow }{ B } \right| =qvBsin\theta \)
(i) F will be maximum, when \(sin\theta =1 \ or \ \theta ={ 90 }^{ o }\)  , i.e., the charged particle is moving perpendicular to the direction of magnetic field.
(ii) F will be minimum, when \(sin\theta =0 \ or \ \theta ={ 0 }^{ o } \ or \ { 180 }^{ o }\) i.e., the charged particle is moving parallel to the direction of magnetic field.

[M¹L¹T^-2A^-2].

The direction of magnetic field is along the +Z-axis, as per right hand rule or Fleming's Left hand rule.

Straight line.

Unit of magnetic field in cgs system is gauss and in SI is tesla or NA^-1m^-1 or weber m^-2. 1 tesla = 10⁴ gauss.

No, a stationary charge does not produce magnetic field.