Physics M.C.Q (1 Marks)

The magnetic field due to the current$-$carrying long, straight wire at point a is given by,
$B =\frac{\mu_0 i }{2 \pi d }$
When both the wires carry currents $i _1$ and $i _2$ in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.
$B ^{\prime}=\frac{\mu_0 i _1}{2 \pi a }-\frac{\mu_0 i _2}{2 \pi a }=10 \mu T \ldots$
Here, $a$ is the distance of the midpoint from both the wires.
When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.
$B ^{\prime \prime}=\frac{\mu_0 i _1}{2 \pi a }+\frac{\mu_0 i _2}{2 \pi a }=30 \mu T \ldots$
On solving eqs. $(1)$ and $(2)$, we get
$i_1-i_2=10$
$i_1+i_2=30$
$\Rightarrow i_1=20$
$i_2=10$
$\frac{i_1}{i_2}=\frac{20}{10}$
$\frac{i_1}{1_2}=\frac{2}{1}$
$\Rightarrow \frac{i_1}{i_2}=2$

The space around a current$-$carrying conductor, in which its magnetic effect can be experienced is called the magnetic field.
When a current is passed through a conductor, it modifies the space around the conductor and forms a magnetic field.

The magnetic field produced by a current $-$ carrying wire at a given point depends directly on the current passing through it.
Therefore, if there is a circular coil having n turns, the field produced is n times as large as that produced by a single turn.
This is because the current in each circular turn has the same direction, and the field due to each turn then just adds up.

The $SI$ unit of the magnetic field is Tesla.

The magnetic force acts in such a way that the direction of the magnetic force and velocity are always perpendicular to each other. If force and velocity are perpendicular force and displacement are also perpendicular, thus $W = FS$ cos $q$, if $q = 90$, work done will be zero.

As per Biot Savart Law, Magnetic field at a point is inversely proportional to square of distance from the current carrying conductor.
Therefore magnitude of magnetic field is same at both points $P$ and $Q,$ irrespective of their position from the conductor.

The magnetic field inside a pipe, i.e. inside a hollow cylindrical wire is zero.
This is due to the symmetry of the situation $($pipe$).$
The pipe can be considered as a series of thin wires arranged in a circle.

Particles $X$ and $Y$ of respective masses $m_1$ and $m_2$ are carrying charge q describing circular paths with respective radii $R_1$ and $R_2$ such that,
$\text{R}_1=\frac{\text{m}_1\text{v}_1}{\text{qB}}$
$\text{R}_1=\frac{\text{m}_2\text{v}_2}{\text{qB}}$
Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.
$\therefore\frac{1}{2}\text{m}_1\text{v}_1^2=\frac{1}{2}\text{m}_2\text{v}_2^2$
$\because\text{R}_1=\frac{\text{m}_1\text{v}_1}{\text{qB}}\Rightarrow\text{v}_1=\frac{\text{R}_1\text{qB}}{\text{m}_1}$
And,
$\text{R}_2=\frac{\text{m}_2\text{v}_2}{\text{qB}}\Rightarrow\text{v}_2=\frac{\text{R}_2\text{qB}}{\text{m}_2}$
$\therefore\text{m}_1\Big(\frac{\text{R}_1\text{qB}}{\text{m}_1}\Big)^2=\text{m}_2\Big(\frac{\text{R}_2\text{qB}}{\text{m}_2}\Big)^2$
$\Rightarrow\frac{\text{m}_1}{\text{m}_2}=\frac{\text{R}_1^2}{\text{R}_2^2}=\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2$

As the current is entering and exiting from two diametrically opposite points of a circular coil, the currents in the two semicircular sections are in the opposite direction.
The fields due to the two semi$-$circular sections at the centre is in the opposite direction.
Hence net feild is zero.

A vertical wire is carrying current in upward direction, so the magnetic field produced will be anticlockwise $($according to the right$-$hand thumb rule$)$. As the electron beam is sent horizontally towards the wire, the direction of the current will be horizontally away from the wire $($direction of conventional current is opposite to the direction of the negative charge$)$. According to Fleming's left$-$hand rule, the force will act in upward direction, deflecting the beam in the same direction.