Practice to calculate the magnetic moment of the coil and on a square loop placed in uniform magnetic field, and Find the torque experienced by dipole in uniform magnetic field and magnetic dipole having dipole moment.
Consider a square loop of side 'a' carrying current \(\text I\), placed in a uniform magnetic field as shown in figure.
Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are perpendicular to each other. So, angle between \(\vec L\) and \(\vec B\) is \(\theta = \) 90º
Force on side 1
\(\vec F _1 = I (\vec L × \vec B)\)
\(= ILB \,sin\,\theta\)
= \(I \,a \, B \,sin \,90º \)
where, \(\theta = \) 90º, L = a
= \(I a \,B \) (into the page)
Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are parallel to each other. So, angle between \(\vec L\) and \(\vec B\) is \(\theta = \) 0º
Force on side 2
\(\vec F _2 = I (\vec L × \vec B)\)
\(= ILB \,sin\,\theta\)
= \(I \,a \, B \,sin \,0º \)
(where, \(\theta = \) 0º, L = a)
= 0
Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are perpendicular to each other. So, angle between \(\vec L\) and \(\vec B\) is \(\theta = \) 90º
Force on side 3
\(\vec F _3 = I (\vec L × \vec B)\)
\(\vec F_3= ILB \,sin\,\theta\)
\(\vec F_3 =I \,a \, B \,sin\, 90º \)
where, \(\theta = \) 90º, L = a
\(\vec F_3 = I\,a\,B \) (outside the page)
Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are parallel to each other. So, angle between \(\vec L\) and \(\vec B\) is \(\theta = \) 0º
Force on side 4
\(\vec F _4 = I (\vec L × \vec B)\)
\(\vec F_4 = ILB \,sin\,\theta\)
\(\vec F_4 =I \,a \, B \,sin\, 0º \)
where, \(\theta = 0°\), L = a
\(\vec F_4 = 0\)
\(\tau = F_1 × \dfrac{a}{2} + F_3 × \dfrac{a}{2} \)
(\(\because\)when net force on a body in zero, torque about every point is same)
\(\tau = IaB \dfrac{a}{2} + IaB \dfrac{a}{2} \)
\(\tau = I a^2 B \)
\(\tau = I AB\) , where A = Area of square plate = a^{2}
Torque \(\vec \tau = IAB \,sin \,\theta\)
\(\vec \tau = I(\vec A × \vec B)\)
where, \(\vec A\) is the area vector.
Note : Using right hand thumb rule , curl fingers in the direction of current, then thumb gives the direction of \(\vec A\) which is inside the page.
\( \vec \mu=I \vec A\)
Then, torque \(\vec \tau = \vec \mu × \vec B\)
Unit:
Unit of magnetic moment \((\vec \mu) = Ampere × m^2\)
\(\vec \mu = N I \, \vec A\)
\(\mu = NIA\)
when only one turn is considered N =1
\(\mu = IA\)
\(\mu = I× \pi r^2\) [\(A = \pi r^2\) ] ............(1)
\(I = \dfrac{q}{T}\)
where, q = charge,T = Time
\(I = \dfrac{q}{\dfrac{2\pi}{\omega}}\) \(\left[\therefore T = \dfrac{2\pi}{\omega}\right]\)
So, \(I = \dfrac{\omega q }{2\pi}\) ............(2)
\(\mu = \dfrac{\omega q}{2\pi} × \pi\, r^2\) \(\left[I = \dfrac{\omega q}{2\pi}\right]\)
\(\mu = \dfrac{\omega\, q\, r^2}{2} \)
A \(12\, \mu A\,m^2\)
B \(3\, \mu A\,m^2\)
C \(6\, \mu A\,m^2\)
D \(8\, \mu A\,m^2\)
\(I = \dfrac {q}{T}\)
\(I = \dfrac {q}{\dfrac{2\pi}{\omega}}\) \( \left[ \therefore T = \dfrac {2\pi}{\omega}\right]\)
or , \(I = \dfrac {q\omega}{{2\pi}} \)
where,
\(\omega=\) angular velocity
q = charge on ring
\(\mu = NIA\) [N=1]
or, \(\mu = IA\)
or, \(\mu = \dfrac{q\omega}{2\pi} A\)
or, \(\mu = \dfrac{q\omega}{2\pi} (\pi R^2)\)
or, \(\mu = \dfrac{\omega R^2 q }{2} \)
\(\vec \tau = \vec \mu × \vec B\)
\(\vec \tau = \dfrac{\omega R^2 qB}{2}\)
A \(8\,\mu\, \text{N-m}\)
B \(6\,\mu\, \text{N-m}\)
C \(3\,\mu\, \text{N-m}\)
D \(2\,\mu\, \text{N-m}\)
Magnetic moment \(\mu _1 = I \vec A\)
where, \(I\) is current
\(\vec A \) is area of loop
\(\mu_1= I \,a^2 (\hat k)\)
Magnetic moment \(\mu _2 = I\vec A \)
\(\mu _2 = I\,a^2 (\hat i)\)
\(\vec \mu = \vec \mu_1+ \vec \mu_2 \)
\(\vec \mu = I\,a^2 (\hat k ) + I\,a^2 (\hat i )\)
\(\vec \mu = I\,a^2 ( \hat i + \hat k )\)
A \(8\,(\hat k + \hat i) \,\text{N-m}^2\)
B \(3 \,(\hat k + \hat i) \,\text{N-m}^2\)
C \(10\,(\hat k + \hat i) \,\text{N-m}^2\)
D \(12 \,(\hat k + \hat i) \,\text{N-m}^2\)
\(\vec \tau = I(\vec A × \vec B)\) where \(\vec A\) is the area vector
Note : Using right hand thumb rule, curl fingers in direction of current, then thumb gives the direction of \(\vec A\) which is inside the page.
\( \vec \mu=I \vec A\)
Then, torque \(\vec \tau = \vec \mu × \vec B\)
Unit:
Unit of magnetic moment, \((\vec \mu) = Ampere × m^2\)
\(\vec \mu = N I \, \vec A\)
Consider a square loop of side 'a' carrying current \(\text I\), placed in a uniform magnetic field as shown in figure.
Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are perpendicular to each other. So, angle between \(\vec L\) and \(\vec B\) is \(\theta = \) 90º
Force on side 1
\(\vec F _1 = I (\vec L × \vec B)\)
= \( ILB \,sin\,\theta\)
= \(I \,a \, B \,sin \,90º \)
where, \(\theta = \) 90º , L = a
= \(I a \,B \) (into the page)
Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are parallel to each other. So, angle between \(\vec L\) and \(\vec B\) is \(\theta = \) 0º
Force on side 2
\(\vec F _2 = I (\vec L × \vec B)\)
\(= ILB \,sin\,\theta\)
= \(I \,a \, B \,sin \,0º \)
(where, \(\theta \)= 0º, L = a)
= 0
Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are perpendicular to each other. So, angle between \(\vec L\) and \(\vec B\) is \(\theta = \) 90º
Force on side 3
\(\vec F _3 = I (\vec L × \vec B)\)
\(\vec F_3= ILB \,sin\,\theta\)
\(\vec F_3 =I \,a \, B \,sin\, 90º \)
where, \(\theta = \) 90º, L = a
\(\vec F_3 = I\,a\,B \) (outside the page)
Since, length (\(\vec L\)) and magnetic field \((\vec B)\) are parallel to each other. So, angle between \(\vec L\) and \(\vec B\) is \(\theta = \) 0º
Force on side 4
\(\vec F _4 = I (\vec L × \vec B)\)
\(\vec F_4 = ILB \,sin\,\theta\)
\(\vec F_4 =I \,a \, B \,sin\, 0º \)
where, \(\theta = \) 0º, L = a
\(\vec F_4 = 0\)
Thus, the magnitude of Torque
\(\tau = F_1 × \dfrac{a}{2} + F_3 × \dfrac{a}{2} \)
(\(\because\)When net force on a body in zero, torque about every point is same.)
\(\tau = IaB \dfrac{a}{2} + IaB \dfrac{a}{2} \)
\(\tau = I a^2 B \)
\(\tau = I AB\)
where, A = Area of square plate = a^{2}
Torque \(\vec \tau = IAB \,sin \,\theta\)
\(\vec \tau = I(\vec A × \vec B)\)
where, \(\vec A\) is the area vector
Note : Using right hand thumb rule, curl fingers in the direction of current, then thumb gives the direction of \(\vec A\) which is inside the page.
\( \vec \mu=I \vec A\)
Then , torque \(\vec \tau = \vec \mu × \vec B\)
Unit:
Unit of magnetic moment \((\vec \mu) = Ampere × m^2\)
\(\vec \mu = N I \, \vec A\)