Answer: Option B. -> √2μ0i3πa⊗  
:
B
According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is half that of ABC i.e. $\frac{i_2}{i_1}=\frac{1}{2}$. Also $i_1+i_2=i\Rightarrow i_1=\frac{2i}{3}$ and $i_2=\frac{i}{3}$
Magnetic field at centre O due to wire AB and BC (part 1 and 2) $B_1=B_2=\frac{{\mu }_0}{4\pi }.\frac{2i_{1}sin{45}^{\circ }}{a/2}\otimes =\frac{{\mu }_0}{4\pi }.\frac{2\sqrt{2}{i}_1}{a}\otimes$
and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) $B_3=B_4=\frac{{\mu }_0}{4\pi }\frac{2\sqrt{2}{I}_2}{a}\odot$
Also $i_1=2i_2$. So $(B_1=B_2)>(B_3=B_4)$
Hence net magnetic field at centre O
$B_{net}=(B_1+B_2)-(B_3+B_4)$
$=2\times \frac{{\mu }_0}{4\pi }.\frac{2\sqrt{2}\times \left(\frac{2}{3}i\right)}{a}-\frac{{\mu }_0}{4\pi }.\frac{2\sqrt{2}\left(\frac{i}{3}\right)\times 2}{a}$
$=\frac{{\mu }_0}{4\pi }.\frac{4\sqrt{2}i}{3a}(2-1)\otimes =\frac{\sqrt{2}{\mu }_{0}i}{3\pi a}\otimes$

Answer: Option A. -> N8
:
A
$\frac{B_c}{M}=\frac{\frac{{\mu }_{0}ni}{2R}}{\pi R^{2}ni}=\frac{{\mu }_0}{2\pi R^3}=NN\alpha \frac{1}{R^3};N^1\alpha \frac{1}{(2R{)}^3}{N}^1\alpha \frac{1}{8R^3}\frac{N^1}{N}=\frac{\frac{1}{8R^3}}{(\frac{1}{R^3}}=\frac{1}{8};N^1=\frac{N}{8}$

Answer: Option A. -> 0,3×10−2T,0
:
A
Inside the core B is finite and rest is zero
$B=\frac{{\mu }_{0}ni}{2\pi r}andr=\frac{r_1+r_2}{2}=\frac{25+26}{2}=25.5=\frac{2\times {10}^{-7}\times 3500\times 11}{25.5\times {10}^{-2}}=\frac{154\times {10}^{-3}}{5-1}=3\times {10}^{-2}T$

Answer: Option A. -> Repulsive force of 10−4N/m  
:
A
By using $\frac{F}{l}=\frac{{\mu }_0}{4\pi }.\frac{2i_1{i}_2}{a}$
$\Rightarrow$ $\frac{F}{l}={10}^{-7}\times \frac{2\times 10\times 5}{0.1}={10}^{-4}N$
Wires are carrying current in opposite direction so the force will be repulsive.

Answer: Option A. -> True
:
A
The magnetic force on a current-carrying wiredepends on the magnitude of current and the strength of magnetic field. The force is maximum when the direction of current and the field are perpendicular to each other.
Here, the direction of magnetic field is acting downwards, the current carrying wire must be kept horizontal for it to be perpendicular to the magnetic field. Also, this force is minimum (i.e. zero), if the current carrying wire and the field are along the same direction or parallel.

Answer: Option D. -> 8
:
D
First cut: 2 north and 2 South Poles.
Second cut: Each of the two pieces were cut into two more pieces.
So, 4 North Poles and 4 South Poles.
Third cut: Each of the 4 pieces is cut into two more pieces.
So, 8 north and 8 South Poles.

Answer: Option A. -> It will not move.
:
A
The direction of the force on aconductor placed in a magnetic field is given by Fleming's left-hand rule. According to this rule, holdingthe forefinger, middle finger, and thumb of the left hand in mutually perpendicular directions, the forefinger indicates the direction of the magnetic field and the middle finger, the direction of the current, then the thumb will indicate the direction of the force which results in the motion of the conductor. However, this force is not experienced when the current flows in the direction of the magnetic field. Here,the direction of the current in the conductor and the magnetic field are the same, so, the conductor will not experience any force. Hence, it will not move in any direction.

Answer: Option B. -> Joule's heating
:
B
Fuseis a protective device used to disconnect the supply when the current flowingthroughit exceeds. It is a wire made up of ametal that has low melting point. High current heats and melts it, breaking the circuit.
Joule's heating is the phenomenon that takes place in the fuse which helps itto break the electrical circuit and hence prevents accidents.

Answer: Option A. -> True 
:
A
In a circuit, when a live wire and a neutral wire come into direct contact (shorted), the current in the circuit increases drastically. The current now flows through a low resistive path consisting of wires rather than domestic electric circuits consisting of resistive components.A fuseis a device which prevents the flow of this large amount of current and thus safeguarding the devices against large current.

Answer: Option A. -> True
:
A
A magnetic field is developed around a current carrying conductor. The magnetic needle is deflected as a result of the interaction of this magnetic field and that around the magnetic needle. Thus, the given statement is true.