Parallel to and above the cable at a distance of 1.5 cm)

7.04 \(\times\) 10⁷ ms^-1

\({ B }_{ v }=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2M\cos { \theta } }{ { r }^{ 3 } } \)
\({ \ B }_{ H }=\frac { { \mu }_{ 0 } }{ 4\pi } \frac { M\sin { \theta } }{ { r }^{ 3 } } \)
\(B=\sqrt { \begin{matrix} { { B }_{ v } }^{ 2 } & + & { { B }_{ H } }^{ 2 } \end{matrix} } =\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } \left[ 4\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right] \)
\(B=\frac { { \mu }_{ 0 }M }{ 4\pi { r }^{ 3 } } \left[ 3\cos ^{ 2 }{ \theta } +1 \right] \)
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) \(\delta ={ 45 }^{ \circ },\cot { \theta =\frac { 1 }{ 2 } } or \ \tan { \theta =2 } \)
\( \theta =\tan ^{ -1 }{ 2is \ the \ locus. } \)

(a) Importance and production of radial magnetic field : In a radial magnetic field magnetic torque ramains maximum for all positions of the coils.
It is produced due to cylindrical pole pieces and soft iron core.
(b) Reason:
Voltmeter: This ensures that a very low current passes through the voltmeter and hence does not change (much) the original potential difference to be measured.
Ammeter: This ensures that the total resistance of the circuit does not change much and the current flowing remains (almost) at its original value.
(c) 

The coil remains suspended in radial magnetic field so that it always experiences maximum torque.
When current passes through the coil, deflection torque \(\tau(\theta)\) is produced given by
\(\tau_{deflection}=NLAB sin 90^{o}\)  --- (i)
As a result, coil rotates and phosphor bronze strip gets twisted. As a result restoring torque given by
\(\tau_{restoring}=k\theta\)   --- (ii)
where, k = torsional restoring constant
∴ In equilibrium,
\(t_{deflecting}=\tau_{restoring}\Rightarrow NIAB=k\theta\)
\(I=(\frac{k}{NAB})\theta \Rightarrow I\propto \theta\)
greater the current, greater the deflection.
(ii) In radial magnetic field, the plane of the coil is always parallel to the plane of the magnetic field and area vector of coil is perpendicular to magnetic field. It is always exerts maximum torque on the coil.
(iii) The voltmeter connected in parallel with the electrical circuit elements to measure potential difference. For exact measurement of PD voltmeter must draw minimum current which is possible only when it has high resistance. Ammeter is connected in series with the electrical circuit and current to be measured passes through it.
In order to protect the galvanometer, a feeble current must pass through the galvanometer, it is possible only when a low resistance (shunt) is connected in parallel with galvanometer to allow the major part of the current to pass through it. 

As the wire carries current I along the y-axis, so \(\overset { \rightarrow }{ l } =l\hat { j } .\) Magnitude force on wire is
\(\overset { \rightarrow }{ F } =I\left( \overset { \rightarrow }{ l } \times \overset { \rightarrow }{ B } \right) =I[l\hat { j } \times [{ B }_{ o }(\hat { i } +\hat { j } +\hat { k } )]]\) 
\(=Il{ B }_{ o }\left[ \hat { j } \times \hat { i } +\hat { j } \times \hat { j } +\hat { j } \times \hat { k } \right] \) 
\(=Il{ B }_{ o }\left[ -\hat { k } +0+\hat { i } \right] =Il{ B }_{ o }\left[ \hat { i } -\hat { k } \right] \) 
Magnitude of the magnetic force is
\(F=Il{ B }_{ o }\left[ { \left( 1 \right) }^{ 2 }+{ \left( -1 \right) }^{ 2 } \right] ^{ 1/2 }=Il{ B }_{ o }\sqrt { 2 } N\)

When a charged particle of charged q moving with velocity v is subjected to normal magnetic field B, then it describe a circular path of radius r given by 
\(Bqv=\frac { m{ v }^{ 2 } }{ r } \ or\ r=\frac { mv }{ Bq } =\frac { p }{ Bq } \) 
where p = momentum of the particle
For the same values of p and B, r\(\propto \) 1/q
\(\therefore \frac { { r }_{ \alpha } }{ { r }_{ p } } =\frac { { q }_{ p } }{ { q }_{ \alpha } } =\frac { e }{ 2e } =\frac { 1 }{ 2 } \)

Inside the pipe, the magnetic field is zero, but there is a finite value of the magnetic field outside the pipe.

Magnetic field induction at the centre of the circular coil carrying current is 
\(B=\frac { { \mu }_{ o } }{ 4\pi } \frac { 2\pi nI }{ r } \ i.e.\ B\propto \frac { n }{ r } \)
\( \therefore \frac { B' }{ B } =\frac { 2n }{ \left( r/2 \right) } \times \frac { r }{ n } =4 \ or \ B'=4B\)

Magnetic field due to current through circular loop is uniform at the centre of the current loop and non-uniform near the circular coil.

(i) The magnetic field of a current element is minimum (which is zero) along the axis of a current element.
(ii) The magnetic field of a current element is maximum in a plane passing through the element and perpendicular to its axis.

Tesla is the SI unit of magnetic field induction or magnetic flux density at a point in the magnetic field. The magnetic field induction at a point in a magnetic field is said to be 1 tesla if one-coulomb charge while moving with a velocity of 1 m/s, perpendicular to the magnetic field experiences a force of 1 N at that point.