Magnetic Effect of Current — Page 5 | JEE Physics

Q41

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is : [Take

\mu

0 = 4

\pi

× 10–7 NA–2]

A 3

\mu

T

B 18

\mu

T

C 9

\mu

T

D 1

\mu

T

Correct Answer

Option B

Solution

For a current carrying wire, magnetic field at a distance r is given by

B=\frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right)

Now, in given case, Due to symmetry of arrangement, net field at centre of triangle is

\begin{aligned} B_{\text{net }} & =\text{ Sum of fields of all wires (sides) } \\\\ & =3 \times \frac{\mu_0 i}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) \end{aligned}

Here, $\theta_1=\theta_2=60^{\circ}$

\begin{aligned} & \therefore \sin \theta_1=\sin \theta_2=\frac{\sqrt{3}}{2}, i=10 \mathrm{~A}, \frac{\mu_0}{4 \pi}=10^{-7} \mathrm{NA}^{-2} \\\\ & \text{ and } r=\frac{1}{3} \times \text{ altitude } \\\\ & \qquad=\frac{1}{3} \times \frac{\sqrt{3}}{2} \times \text{ sides length }=\frac{1}{2 \sqrt{3}} \times 1 \mathrm{~m}=\frac{1}{2 \sqrt{3}} \mathrm{~m} \end{aligned}

So,

\begin{aligned} B_{\mathrm{net}} & =\frac{3 \times 10^{-7} \times 10 \times 2\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{2 \sqrt{3}}\right)}=18 \times 10^{-6} \mathrm{~T} \\\\ \Rightarrow B_{\mathrm{net}} & =18 \mu \mathrm{T} \end{aligned}

Q42

A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion constant 10–6 N-m/rad. The coil is placed in a maganetic field B parallel to its plane. The coil deflects by 1° for a current of 1 mA. The value of B (in Tesla) is approximately :-

A 10–4

B 10–2

C 10–1

D 10–3

Correct Answer

Option D

Solution

NIAB = KQ 175 × 1 × 10–3 × 1 × 10–4 × B =

{{{{10}^{ - 6}} \times \pi } \over {180}}

\Rightarrow B = {\pi \over {180}} \times {{10} \over {175}} \approx 9.97 \times {10^{ - 4}}\,T

$\Rightarrow$ B = 10–3 T

Q43

A proton and an

\alpha

-particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp : r

\alpha

of the circular paths described by them will be ;

A

1:\sqrt 3

B 1 : 3

C

1:\sqrt 2

D 1 : 2

Correct Answer

Option C

Solution

KE = q

\Delta

V r =

{{\sqrt {2mq\Delta V} } \over {qB}}

r $\propto$

\sqrt {{m \over q}}

{{{r_p}} \over {{r_ \propto }}}

=

{1 \over {\sqrt 2 }}

Q44

A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field B

{\mathop i\limits^ \wedge }

. The torque on the coil due to the magnetic field is :

A

{{B{r^2}I} \over {\pi N}}

B B

\pi

r2IN

C Zero

D

{{B\pi{r^2}I} \over { N}}

Correct Answer

Option B

Solution

According to the question, the situation can be drawn as Let the current I is flowing in anti-clockwise direction, then the magnetic moment of the coil is m = NIA where, N = number of turns in coil and A = area of each coil = $\pi$ r2.

Its direction is perpendicular to the area of coil and is along Y-axis.

Then, torque on the current coil is

\tau = m \times B = mB\sin 90^\circ = NIAB

= NI\pi {r^2}B(N - m)

Q45

Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm, 50 turns and carrying current I (Ampere) in units of

{{{\mu _0}I} \over \pi }

is :

A 250

\sqrt 3

B 5

\sqrt 3

C 500

\sqrt 3

D 50

\sqrt 3

Correct Answer

Option C

Solution

\tan 30 = {x \over d}

d = {x \over {\tan 30}}

d = {{5 \times {{10}^{ - 2}}} \over {{1 \over {\sqrt 3 }}}}

d = 5\sqrt 3 \times {10^{ - 2}}

For one part of the wire with N turns,

B = {{{\mu _0}IN} \over {4\pi d}}(\sin {\theta _1} + \sin {\theta _2})

For 6 identical parts of the wire,

{B_{net}} = 6B

= {{6{\mu _0}IN} \over {4\pi d}}(\sin 30 + \sin 30)

= {{{\mu _0}I} \over \pi }\left( {{{6 \times 50} \over {4 \times 5\sqrt 3 \times {{10}^{ - 2}}}}} \right)\left( {2 \times {1 \over 2}} \right)

= 500\sqrt 3 \left( {{{{\mu _0}I} \over \pi }} \right)

Q46

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then -

A Th = 1.5 Tc

B Th = Tc

C Th = 2Tc

D Th = 0.5 Tc

Correct Answer

Option B

Solution

T =

2\pi \sqrt {{1 \over {\mu B}}}

Th =

2\pi \sqrt {{{m{R^2}} \over {\left( {2\mu } \right)B}}}

TC =

2\pi \sqrt {{{1/2m{R^2}} \over {\mu B}}}

Q47

The region between y = 0 and y = d contains a magnetic field

\overrightarrow B = B\widehat z

. A particle of mass m and charge q enters the region with a velocity

\overrightarrow v = v\widehat i.

If d

=

{{mv} \over {2qB}},

the acceleration of the charged particle at the point of its emergence at the other side is :

A

{{qvB} \over m}\left( -{{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)

B

{{qvB} \over m}\left( {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right)

C

{{qvB} \over m}\left( {{{ - \widehat j + \widehat i} \over {\sqrt 2 }}} \right)

D

{{qvB} \over m}\left( {{{\widehat j + \widehat i} \over {\sqrt 2 }}} \right)

Correct Answer

Option A

Solution

Here R =

{{mv} \over {qB}}

= 2d cos $\theta$ =

{{{R \over 2}} \over R}

=

{1 \over 2}

$\Rightarrow$ $\theta$ = 60o Acceleration of the charged particle at the point of its emergence,

\overrightarrow {{a_c}} = {a_{{c_x}}}\left( { - \widehat i} \right) + {a_{{c_y}}}\left( { - \widehat j} \right)

=

{a_c}\cos 30^\circ \left( { - \widehat i} \right) + {a_c}\sin 30^\circ \left( { - \widehat j} \right)

=

{a_c}\left( {{{\sqrt 3 } \over 2}\left( { - \widehat i} \right) + {1 \over 2}\left( { - \widehat j} \right)} \right)

=

{{qvB} \over m}\left( { - {{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)

Q48

A particle of mass m and charge q is in an electric and magnetic field given by

\overrightarrow E = 2\widehat i + 3\widehat j;\,\,\,\overrightarrow B = 4\widehat j + 6\widehat k.

The charged particle is shifted from he origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is :

A (2.5) q

B (0.35) q

C (0.15) q

D 5 q

Correct Answer

Option D

Solution

{\overrightarrow F _{net}} = q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)

= \left( {2q\widehat i + 3q\widehat j} \right) + q\left( {\overrightarrow v \times \overrightarrow B } \right)

W = {\overrightarrow F _{net}}.\overrightarrow S

$=$ 2q + 3q $=$ 5q

Q49

A loop of flexible wire of irregular shape carrying current is placed in an external magnetic field. Identify the effect of the field on the wire.

A Loop assumes circular shape with its plane normal to the field.

B Loop assumes circular shape with its plane parallel to the field.

C Wire gets stretched to become straight.

D Shape of the loop remains unchanged.

Correct Answer

Option A

Solution

Force on each wire be along radially outward and equal so, it will take the shape of circle and parallel to the field.

Q50

A square loop of side 2

a

and carrying current I is kept in xz plane with its centre at origin. A long wire carrying the same current I is placed parallel to z-axis and passing through point (0, b, 0), (b >> a). The magnitude of torque on the loop about z-axis will be :

A

{{2{\mu _0}{I^2}{a^2}} \over {\pi b}}

B

{{2{\mu _0}{I^2}{a^2}b} \over {\pi \left( {{a^2} + {b^2}} \right)}}

C

{{{\mu _0}{I^2}{a^2}b} \over {2\pi \left( {{a^2} + {b^2}} \right)}}

D

{{{\mu _0}{I^2}{a^2}} \over {2\pi b}}

Correct Answer

Option A

Solution

First, let's consider the magnetic field created by the long wire carrying a current $I$ at a distance $b$ .

According to Ampere's Law, the magnetic field $B$ at a distance $b$ from a long straight wire is given by:

B = \frac{{\mu_0 I}}{{2\pi b}}

where $\mu_0$ is the permeability of free space.

Given that the square loop of side $2a$ is carrying the same current $I$ and is located in the xz-plane with its center at the origin, we can analyze the forces on each side of the loop.

The sides of the loop parallel to the x-axis will experience forces due to the magnetic field from the long wire.

Considering symmetry and the directions of forces, the net force on these sides will not contribute to the torque around the z-axis.

The contribution to the torque around the z-axis will predominantly come from the sides of the loop parallel to the y-axis.

For these sides, the magnetic forces will be in opposite directions and will create a torque around the z-axis.

Let's calculate the forces on the sides parallel to the y-axis.

For a current element $Idl$ in the presence of a magnetic field $B$ , the force $dF$ is given by:

dF = I dl \times B

For the sides at $x = a$ and $x = -a$ , the distances to the wire are $a + b$ and $a - b$ , respectively.

The magnetic fields at these positions due to the long wire are: For $x = a$ :

B_a = \frac{{\mu_0 I}}{{2 \pi (a+b)}}

For $x = -a$ :

B_{-a} = \frac{{\mu_0 I}}{{2 \pi (a-b)}}

Since $b \gg a$ , we can approximate these fields using binomial expansion for small $\left(\dfrac{a}{b}\right)$ :

B_a \approx \frac{{\mu_0 I}}{{2 \pi b}} \left(1 - \frac{a}{b}\right)

B_{-a} \approx \frac{{\mu_0 I}}{{2 \pi b}} \left(1 + \frac{a}{b}\right)

The forces on each side of the loop with length $2a$ are: For $x = a$ :

F_a = I \cdot 2a \cdot B_a = I \cdot 2a \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left(1 - \frac{a}{b}\right)

For $x = -a$ :

F_{-a} = I \cdot 2a \cdot B_{-a} = I \cdot 2a \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left(1 + \frac{a}{b}\right)

The net torque $\tau$ around the z-axis is due to these forces, with lever arms $a$ and $-a$ respectively:

\tau = 2a \left( F_a - F_{-a} \right)

Substituting the expressions for $F_a$ and $F_{-a}$ :

\tau = 2a \left[ I \cdot 2a \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left(1 - \frac{a}{b}\right) - I \cdot 2a \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left(1 + \frac{a}{b}\right) \right]

Simplifying this expression, we get:

\tau = 2a \left[ 2a I \cdot \frac{{\mu_0 I}}{{2 \pi b}} \left( -\frac{2a}{b} \right) \right]

\tau = -2a \cdot \frac{{4a^2 \mu_0 I^2}}{{2 \pi b^2}}

The negative sign indicates the direction of the torque, but the magnitude is:

\tau = \frac{{4a^3 \mu_0 I^2}}{{\pi b^2}}

Since $a \ll b$ , the approximate magnitude of the torque around the z-axis simplifies to:

\tau = \frac{{2 \mu_0 I^2 a^2}}{{\pi b}}

Therefore, the correct answer is: Option A

\frac{{2\mu_0 I^2 a^2}}{{\pi b}}