
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is \[{60^ \circ }\], and one of the fields has a magnitude of \[1.2 \times {10^{ - 2}}T\]. If the dipole comes to stable equilibrium at an angle of \[{15^ \circ }\] with this field, what is the magnitude of the field?
A \[3\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}T\]
B \[\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}T\]
C \[6\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}T\]
D \[2\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}T\]
Answer
496.8k+ views
Hint: When an electric current flows through a wire, a magnetic field can be created around it. Around the wire, concentric circles the magnetic field lines formed. The direction of the magnetic field depends on the direction of the current. It can be resolved by using the right-hand rule. The magnitude of the magnetic field depends on the distance from the charge-carrying wire and the amount of current. The formula includes the constant, that is the permeability of free space.
Formula used:
Magnetic field magnitude \[B = \dfrac{{{\mu _0}I}}{{2\pi r}}\], \[{\mu _0} = \] the permeability of free space.
\[M{B_1}\sin {\theta _1} = M{B_2}\sin {\theta _2}\]
Complete answer:
Let, Magnitude of one of the magnetic fields, \[{B_1} = 1.2 \times {10^{ - 2}}T\]
The magnitude of the other magnetic field \[{B_2}\]
The angle between the two fields, \[\theta = {60^ \circ }\].
A stable equilibrium angle between dipole and field \[{B_1}\]\[{\theta _1} = {15^ \circ }\] .
The angle between dipole and field \[{B_2}\], \[{\theta _2} = \theta - {\theta _1} = {60^ \circ } - {15^ \circ } = {45^ \circ }\]
The torque between both the poles balances each other.
The torque due to field \[{B_1}\]\[ = \] Torque due to field \[{B_2}\]
\[M{B_1}\sin {\theta _1} = M{B_2}\sin {\theta _2}\] (\[M = \] Magnetic moment of the dipole)
Therefore, \[{B_2} = \dfrac{{{B_1}\sin {\theta _1}}}{{\sin {\theta _2}}}\]
\[ \Rightarrow {B_2} = 1.2 \times {10^{ - 2}} \times \dfrac{{\sin {{15}^ \circ }}}{{\sin 45 \circ }}\]
\[ \Rightarrow {B_2} = 4.39 \times {10^{ - 3}}T\]
\[ \Rightarrow {B_2} \approx 3\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}T\]
Hence, the correct answer is option A.
Note:
The unit of the magnetic field is Tesla.
The magnetic dipole moment is defined in terms of the torque in a magnetic field. In The same magnetic field, the larger torque creates the larger magnetic moment. The torque doesn't only depend on the magnitude of the magnetic moment but also the orientation relative to the magnetic field.
Formula used:
Magnetic field magnitude \[B = \dfrac{{{\mu _0}I}}{{2\pi r}}\], \[{\mu _0} = \] the permeability of free space.
\[M{B_1}\sin {\theta _1} = M{B_2}\sin {\theta _2}\]
Complete answer:
Let, Magnitude of one of the magnetic fields, \[{B_1} = 1.2 \times {10^{ - 2}}T\]
The magnitude of the other magnetic field \[{B_2}\]
The angle between the two fields, \[\theta = {60^ \circ }\].
A stable equilibrium angle between dipole and field \[{B_1}\]\[{\theta _1} = {15^ \circ }\] .
The angle between dipole and field \[{B_2}\], \[{\theta _2} = \theta - {\theta _1} = {60^ \circ } - {15^ \circ } = {45^ \circ }\]
The torque between both the poles balances each other.
The torque due to field \[{B_1}\]\[ = \] Torque due to field \[{B_2}\]
\[M{B_1}\sin {\theta _1} = M{B_2}\sin {\theta _2}\] (\[M = \] Magnetic moment of the dipole)
Therefore, \[{B_2} = \dfrac{{{B_1}\sin {\theta _1}}}{{\sin {\theta _2}}}\]
\[ \Rightarrow {B_2} = 1.2 \times {10^{ - 2}} \times \dfrac{{\sin {{15}^ \circ }}}{{\sin 45 \circ }}\]
\[ \Rightarrow {B_2} = 4.39 \times {10^{ - 3}}T\]
\[ \Rightarrow {B_2} \approx 3\left( {\sqrt 3 - 1} \right) \times {10^{ - 3}}T\]
Hence, the correct answer is option A.
Note:
The unit of the magnetic field is Tesla.
The magnetic dipole moment is defined in terms of the torque in a magnetic field. In The same magnetic field, the larger torque creates the larger magnetic moment. The torque doesn't only depend on the magnitude of the magnetic moment but also the orientation relative to the magnetic field.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

