Mix Test - 6

Given: Initially force per unit length on the wires \(= f _{ ab }= f _{ ba }= f =10^{-3}\) N and \(d =2\) m

If two current carrying long wires \(A\) and \(B\) are separated by a very small distance and placed parallel to each other. Then the force per unit length on wire \(A\) and wire \(B\) is given as,

\(f_{a b}=f_{b a}=f=\frac{\mu_{o} I_{a} I_{b}}{2 \pi d}\)

Where, \(I _{ a }=\) current in the wire \(A , I _{ b }=\) current in the wire \(B\), and \(d =\) distance between the wires

If the current in both the wires is doubled,

\(I_{a}^{\prime}=2 I_{a}\)

\(\Rightarrow I_{b}^{\prime}=2 I_{b}\)

And the distance between the wires is halved, so,

\(d^{\prime}=\frac{d}{2}\)

So, when the current in both the wires is doubled and the distance between the wires is halved, the new force per unit length on wire A and wire B will be,

\(f_{a b}^{\prime}=f_{b a}^{\prime}=f^{\prime}=\frac{\mu_{o} I_{a}^{\prime} I_{b}^{\prime}}{2 \pi d^{\prime}}\)

\(\Rightarrow f^{\prime}=2 \times \frac{\mu_{o} \times 2 I_{a} \times 2 I_{b}}{2 \pi d}\)

\(\Rightarrow f^{\prime}=8 \times \frac{\mu_{o} I_{a} I_{b}}{2 \pi d}\)

\(\Rightarrow f ^{\prime}=8 f\)

\(\Rightarrow f ^{\prime}=8 \times 10^{-3}\) N

Hence, the correct option is (C).

When the velocity of a charged particle is perpendicular to a magnetic field, it describes a circle and the radius of the circle is given by:

\(r=\frac{m v}{q B} \quad \ldots\) (1)

Where, \(m=\) mass, \(q=\) magnitude of charge, \(v=\) speed of charge, \(B=\) magnetic field 

We know that \(q , m\), and \(B\) remain unchanged during the motion of a charged particle.

\(r \propto v \quad \ldots\) (2)

\(v _{1}> v _{2}\)

\(r _{1}> r _{2}\)

Hence, the correct option is (B).

Coercivity is defined as the minimum value of magnetising intensity that is required to bring the material to its original state. In the case of a ferromagnetic substance, when the applied magnetic field is increased in the reverse direction, then at some point H_c the substance completely loses its magnetization.

Hence, the correct option is (B).

When the coil deflects, the suspension wire is twisted. On account of elasticity, a restoring couple is set up in the wire. This couple is proportional to the twist.

If \(\theta=\) the angular twist, then,

The moment of the restoring couple \(= C \theta\)

Where \(C=\) restoring couple per unit twist.

At equilibrium,

Deflecting couple \(=\) Restoring couple

So, we can write,

\(nBIA = C \theta\)

\(\Rightarrow \theta=\left(\frac{n B A}{C}\right) I\)

Here, \(n, B\), and \(A\) are constant.

Therefore, the deflection produced in the moving coil galvanometer is directly proportional to the amount of current 'I' passing through it.

Thus, if the current flowing through the moving coil galvanometer is increased, then the deflection in the coil will also increase.

Hence, the correct option is (A).

Given:

\(q =1.6 \times 10^{-19}\) C,

\(m =9.1 \times 10^{-31}\) kg

\(B =2 \times 10^{-4}\) weber/m\(^{2}\)

The time period is given as,

\(T=\frac{2 \pi m}{q B}\)

Where, \(m=\) mass, \(q=\) magnitude of charge, \(B=\) magnetic field

\(T=\frac{2 \pi \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 2 \times 10^{-4}}\)

\(\Rightarrow T=1.79 \times 10^{-7}\) sec

Hence, the correct option is (A).

Given: Radius of the conductor \(r=2\) mm, current through the conductor \(i=5\) A, the magnetic field at distance \(d\) from the center \(=1\) mT, here d \(<\) r

According to Ampere's law

\(\oint B \cdot d l=\mu_{0}i\)

The magnetic field is always parallel to the element \(dl\) on the Ampere loop,

\(B .2 \pi d=\mu_{0} i\)

Where, \(i\) is the current inside the loop

\(d=\frac{\mu_{0} i}{2 \pi B}\)

\(\Rightarrow d=\frac{4 \pi \times 10^{-7} \times 5}{2 \pi \times 1 \times 10^{-3}}=1 \times 10^{-3}=1\) mm

Hence, the correct option is (C).

The magnetic field B inside the toroid is constant in magnitude for the ideal toroid of closely wound turns. The magnetic field in the open space inside (point P) and exterior to the toroid (point Q) is zero.

The magnetic field \(B\) inside the toroid is given as,

\(B=\frac{\mu_{o} N I}{2 \pi r}\)

Where, \(N =\) number of turns, \(I =\) current, and \(r =\) average radius of the toroid, If \(N\) and \(I\) are constant, then,

\(B \propto \frac{1}{r}\)

\(\Rightarrow B \propto r^{-1}\)

Hence, the correct option is (C).

We know that when a particle enters the magnetic field in a direction perpendicular to the direction of the field, it undergoes circular motion.

The magnetic force is then equivalent to the centripetal force and is given as:

\(F=\frac{m v^{2}}{r}=q v B\)

\(\Rightarrow r=\frac{m v}{q B}\)

So, the radius of curvature \(r=\frac{m v}{q B}\)

Where, \(m=\) mass, \(q=\) magnitude of charge, \(v=\) speed of charge, \(B=\) magnetic field 

It is given that the particles have the same charge (q) and initial velocity (v). Also, the magnetic field (B) is uniform.

Thus, r ∝ m

​Therefore, the radius of curvature is greater for the particle that has more mass.

Hence, the correct option is (B).

Given that:

Current sensitivity \(\left( S _{ i }\right)=5\) div/(mA) \(=5 \times 10^{3}\) div/A

Voltage sensitivity \(( S_{v} )=20\) div/V

We know that:

Voltage sensitivity \(\left( S _{ v }\right)=\frac{S _{i}}{R}\)

Resistance \(( R )= \frac{S_i}{S_{v}}=\frac{(5 \times 10^{3} \operatorname{div}/A)}{(20 \operatorname{div}/V)}=250 \Omega\)

Hence, the correct option is (C).

Biot-Savart Law: The law that gives the magnetic field generated by a constant electric current is Biot-savart law.

Let us take a current-carrying wire of current \(I\) and we need to find the magnetic field at a distance \(r\) from the wire then it is given by:

\(d B=\frac{\mu_{0} I}{4 \pi}\left(\frac{\overrightarrow{d l} \times \hat{r}}{r^{2}}\right)\)

Where, \(\mu_{0}=4 \pi \times 10^{-7}\) T.m/A is the permeability of free space/vacuum, \(dl =\) small element of wire and \(\hat{r}\) is the unit position vector of the point where we need to find the magnetic field.

From the above expression of the Biot-savart law, the magnetic field is: 

• Directly proportional to the length of the wire. So statement (i) is correct.
• Directly proportional to the electric current. So statement (iii) is correct.
• Biot-savart law gives the magnetic field, not the electric field. So statement (ii) is wrong.

Hence, the correct option is (C).

When a soft iron core is inserted inside the solenoid then the strength of the magnetic field becomes very large because the iron core gets magnetized by induction.

The soft iron core helps in concentrating the magnetic lines of forces through the solenoid so that the magnetic field is almost uniform at the end face of the core.

Thus, the strength of the magnetic field increases when a soft iron core is inserted inside a solenoid.

Hence, the correct option is (A).

Given: Radius of Coil \(A=a\)

Current in coil \(A=1\)

Field due to Coil A at center,

\(B_{A}=\frac{\mu_{0} I}{2a} \quad \ldots\) (i)

Radius of Coil \(B=2 a\)

Current in coil \(B=1\)

Field due to Coil B at center,

\(B_{B}=\frac{\mu_{0} I }{2(2 a )}=\frac{1}{2}\left(\frac{\mu_{0} I }{2 a }\right)\quad \ldots\) (ii)

Comparing (i) and (ii),

\(B_{B}=\frac{B_{A}}{2}\)

\(\Rightarrow B _{ A }: B _{ B }=2: 1\)

So, the required ratio is \(2: 1\).

Hence, the correct option is (A).

If a rectangular loop carrying a steady current is placed in a uniform magnetic field then it will experience a torque. The net force on the loop will be zero.

The torque on the current-carrying rectangular loop is given as,

τ = NIAB sinθ

Where, N = number of turns in the coil, I = current in the loop, A = area enclosed by the loop, B = magnetic field intensity, and θ = angle between the normal to the plane of the coil and the direction of a uniform magnetic field.

Hence, the correct option is (A).

Magnetic field due to Straight conductor at distance r is:

\(\mathrm{B}=0.4\)

\(\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}=0.4 \text { Tesla } \quad \ldots(1)\)

Now, μ₀ and i is constant for now, as the only distance is changing to 2r.

So, we can say that:

\(\mathrm{B} \propto \frac{1}{\mathrm{r}}\)

or

\(\mathrm{B}=\mathrm{k} \frac{1}{\mathrm{r}} \quad \ldots(2)\)

\(\mathrm{k}\) is constant.

Now, if the radius is increased to \(2 \mathrm{r}\), the new field will be \(\mathrm{B}^{\prime}\)

\(\mathrm{B}^{\prime}=\mathrm{k} \frac{1}{2 \mathrm{r}} \quad \ldots(3)\)

Using (2) and (3)

\(\mathrm{B}^{\prime}=\frac{\mathrm{B}}{2} \quad \cdots(4)\)

Using (1) and (4)

\(\mathrm{B}^{\prime}=\frac{0.4}{2} \text { Tesla }\)

\( \mathrm{B}^{\prime}=0.2 \text { Tesla }\)

So, the new magnetic field strength will be 0.2 Tesla.

Hence, the correct option is (A).

From above it is clear that when a current-carrying rectangular coil is placed in a magnetic field it experiences a torque.

\(\therefore\) The moment of the deflecting couple \((\tau)=n B i A\)

Where, \(n =\) number of turns, \(B =\) magnetic field, \(i =\) current and \(A =\) area of coil

When the coil deflects, the suspension wire is twisted. On account of elasticity, a restoring couple is set up in the wire. This couple is proportional to the twist.

If \(\theta=\) the angular twist, then,

The moment of the restoring couple \(=C \theta\)

Where, \(C=\) restoring couple per unit twist.

At equilibrium,

Deflecting couple \(=\) Restoring couple

So, we can write,

\(nBi A = C \theta\)

\(\Rightarrow\left(\frac{n B A}{C}\right) i=\theta\)

Here, \(n , B\), and \(A\) are constant

Therefore, the deflection produced in the moving coil galvanometer is directly proportional to the amount of current passing through it.

Hence, the correct option is (B).

Given:

\(L=40\) cm \(=0.4\) m

\(W =10\) cm \(=0.1\) m

\(N =10\) turns

\(I =16\) A

\(\theta=60^{\circ}\)

\(B =0.60\) T

Where, \(N=\) number of turns in the coil, I = current in the loop, \(A=\) area enclosed by the loop, \(L=\) length of the loop, \(W=\) width of the loop, \(B=\) magnetic field intensity, and \(\theta=\) angle between the normal to the plane of the coil and the direction of a uniform magnetic field.

The area enclosed by the loop is given as,

\(A = L \times W\)

\(\Rightarrow A =0.4 \times 0.1\)

\(\Rightarrow A =0.04\) m\(^{2}\)

So, the torque on the current-carrying rectangular loop is given as,

\(\tau =IAB \sin \theta\)

\(\Rightarrow \tau =16 \times 0.04 \times 0.6 \times \sin 60\)

\(\Rightarrow \tau=\frac{16 \times 4 \times 6 \times \sqrt{3}}{10 \times 10 \times 2}\)

\(\Rightarrow \tau=1.92 \sqrt{3}\) N-m

Hence, the correct option is (B).

If N= number of turns in the coil, I= current in the loop, A= area enclosed by the loop, B= magnetic field intensity, and θ= angle between the normal to the plane of the coil and the direction of a uniform magnetic field.

Then the torque on the current-carrying rectangular loop is given as,

T=NIAB⋅sinθ…(1)

If N,A,I, and B is constant, then,

T∝sinθ…(2)

We know that the minimum and the maximum angle possible between the area vector and the magnetic field is 0∘ and 90∘.

We know that when the value of θ is increased from 0∘ to 90∘, the value of sinθ also increases.

Therefore, if the angle between the area vector and the magnetic field is increased then the value of sinθ increases and by equation (2) we can say that the torque on the loop will also increase.

Hence, the correct option is (A).

The magnetic field \(B\) inside the toroid is given as,

\(B=\frac{\mu_{o} N I}{2 \pi r}\quad \ldots\) (1)

Where, \(N =\) number of turns, \(I=\) current, and \(r =\) average radius of the toroid

When the current in the toroid is doubled, then:

\(I^{\prime}=2I\quad \ldots\) (2)

So, the magnetic field in the toroid when the current is doubled is given as,

\(B^{\prime}=\frac{\mu_{o} N I^{\prime}}{2 \pi r}\)

\(\Rightarrow B^{\prime}=\frac{2 \mu_{o} N I}{2 \pi r}\quad \ldots\) (3)

By equation (1) and equation (3),

\(B^{\prime}=2B\)

Hence, the correct option is (A).

Biot-Savart’s law is an equation that gives the magnetic field produced due to a current-carrying segment. Biot-Savart’s law is used to determine the magnetic field at any point due to a current-carrying conductor. This law is although for infinitesimally small conductors yet it can be used for long conductors.

According to this law, the magnetic field produced at a distance r from the current-carrying element is given as,

\(\overrightarrow{d B}=\frac{\mu_{o}}{4 \pi} \frac{\overrightarrow{d l} \times \vec{r}}{r^{3}}\)

Where, \(\overrightarrow{d B}=\) magnetic field, \(r =\) distance of a point from the element, \(dl =\) length of the element, and \(\theta=\) angle between \(d l\) and \(r\)

Hence, the correct option is (D).

When the current flows in the same direction in the two parallel wires then both wires attract each other and if the current flows in the opposite direction in the two parallel wires then both wires repel each other.

Therefore, if the direction of flow of current is opposite in the two parallel wires separated by a small distance, then the wires will repel each other.

Hence, the correct option is (B).

Given:

q= charge, m= mass, B= magnetic field, v= velocity, and θ=90∘

So, the force on the charged particle due to the magnetic field is given as,

F=qvBsinθ

⇒F=qvB×sin90∘

⇒F=qvB⋯(1)

By Newton's second law of motion, the force is given as,

F=ma⋯(2)

Where, a= acceleration

By equation (1) and equation (2),

ma=qvB

⇒a=qvBm

Hence, the correct option is (A).

The permanent magnets can be made in the following ways:

1. By holding an iron rod in the north-south direction and hammer it repeatedly.
2. By holding a steel rod and stroke it with one end of a bar magnet a large number of times.
3. An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass a current.

Hence, the correct option is (C).

A moving coil galvanometer can be converted into a ammeter by connecting a low resistance in parallel to the moving coil galvanometer.

A galvanometer can be converted into an ammeter by connecting a shunt resistance in parallel to it. The shunt resistance should have very low resistance. So, the ammeter (the parallel combination of galvanometer and shunt resistance) will have low resistance.

Hence, the correct option is (B).

Electromagnets are used in electric bells, loudspeakers,

• An electromagnet is a type of magnet in which the magnetic field is produced by an electric current.
• Electromagnets usually consist of wire wounds into a coil.
• The magnetism of the electromagnets can be increased by placing a soft iron rod inside the solenoid.
• The core of electromagnets is made of ferromagnetic materials which have high permeability and low retentivity.
• Electromagnets are used in electric bells, loudspeakers, and telephone diaphragms.
• Giant electromagnets are used in cranes to lift machinery, and bulk quantities of iron and steel.

Hence, the correct option is (C).

Given: \(N=30\) turns, \(x=20\) cm \(=0.2\) m, and \(I=10\) A

Where, \(N=\) number of turns in the coil, \(I=\) current in the loop, \(x=\) side of the square loop, and \(A =\) area enclosed by the loop

The area enclosed by the square loop is given as,

\(A = x ^{2}\)

\(\Rightarrow A =0.2^{2}\)

\(\Rightarrow A =0.04\) m \(^{2}\)

We know that the magnetic moment '\(m\)' of the current-carrying loop is given as,

\(m = NIA\)

\(\Rightarrow m =30 \times 10 \times 0.04\)

\(\Rightarrow m =12\) A-m\(^{2}\)

Hence, the correct option is (A).

If \(N =\) number of turns in the coil, \(I =\) current in the loop, \(A =\) area enclosed by the loop, \(B=\) magnetic field intensity, and \(\theta=\) angle between the normal to the plane of the coil and the direction of a uniform magnetic field.

Then the torque on the current-carrying rectangular loop is given as,

\(T = NIAB \cdot \sin \theta \quad \ldots\)(1)

By equation (1) it is clear that the torque will be maximum when the value of \(\sin \theta\) is maximum.

We know that the maximum value of \(\sin \theta\) is 1.

For \(\sin \theta=1\)

\(\Rightarrow \theta=90^{\circ}\)

So, the torque on the loop will be maximum when the angle between the area vector and the magnetic field is \(90^{\circ}\).

Hence, the correct option is (B).

The magnetism of the electromagnets can be increased by placing a soft iron rod inside the solenoid.

• The core of electromagnets is made of ferromagnetic materials which have high permeability and low retentivity.
• An electromagnet is a type of magnet in which the magnetic field is produced by an electric current.
• Electromagnets usually consist of wire wounds into a coil.
• The magnetism of the electromagnets can be increased by placing a soft iron rod inside the solenoid.
• The core of electromagnets is made of ferromagnetic materials which have high permeability and low retentivity.
• Electromagnets are used in electric bells, loudspeakers, and telephone diaphragms.
• Giant electromagnets are used in cranes to lift machinery, and bulk quantities of iron and steel.

Hence, the correct option is (A).

We know that the voltage sensitivity is given as,

\(V_{s}=\left(\frac{I_{s}}{R}\right) \quad \ldots\)(1)

Where, \(V _{ s }=\) voltage sensitivity, \(I _{ s }=\) current sensitivity, and \(R =\) resistance

When the resistance is increased by factor 2,

\(R^{\prime}=2 R\quad \ldots\)(2)

\(I_{s}^{\prime}=I_{s}+20 \% \text { of } I_{s}\)

\(\Rightarrow I_{s}^{\prime}=I_{s}+\frac{20}{100} I_{s}\)

\(\Rightarrow I_{s}^{\prime}=1.2 I_{s}\quad \ldots\)(3)

From equation (1) we get,

\( V_{s}^{\prime}=\frac{I_{s}^{\prime}}{R^{\prime}}\)

\(\Rightarrow V_{s}^{\prime}=\frac{1.2 I_{s}}{2 R}\quad \ldots\)(4)

By equation (1) and equation (4)

\(V_{s}^{\prime}=0.6 V_{s}\quad \ldots\)(5)

The change in voltage sensitivity is given as,

\(\% \Delta V=\frac{V_{s}^{\prime}-V_{s}}{V_{s}} \times 100\)

\(\Rightarrow \% \Delta V=\frac{0.6 V_{s}-V_{s}}{V_{s}} \times 100\)

\(\Rightarrow \% \Delta V=-40 \%\)

Hence, the correct option is (B).

Given:

\(B=7 \pi \times 10^{-5}\) Wb/m\(^{2}\) and loop of radius \(=5\) cm \(=5 \times 10^{-2}\) m

The magnetic field at the center of the circular coil is given by,

\(B=\frac{\mu_{o}}{2} \frac{I}{r}\)

\(\Rightarrow I=\frac{2 B r}{\mu_{o}}\)

Where, \(B =\) strength of the magnetic field, \(I =\) current, \(r =\) radius or distance

Substitute the value of \(B, r\), and \(\mu_{0}\) in the above equation, we get,

\(I=\frac{2 \times 7 \pi \times 10^{-5} \times 5 \times 10^{-2}}{4 \pi \times 10^{-7}}=17.5\) A

Hence, the correct option is (B).

Ampere's law

The circulation \(\oint B.dl\) of the resultant magnetic field along a closed, plane curve is equal to \(\mu_{0}\) times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant.

\(\oint B.dl =\mu_{0} \)i

Where, \(B\) is the magnetic field at the length element \(dl\) of the loop. i is the sum of all currents inside the loop.

The magnetic field B on the ampere loop element \(dl\) is due to all the current elements in the vicinity. But on the right-hand side of the equation, 'i' is the sum of all the currents inside the loop only.

Hence, the correct option is (B).