Force And Torque Due To Magnetic Field L-1 Force and Torque due to Magnetic Field
→ \rightarrow → → \rightarrow → Force and Torque due to Magnetic Field → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Force on a Current Carrying Conductor
F ⃗ B = q ( v ⃗ × B ⃗ ) \vec{F}_B=q(\vec{v} \times \vec{B}) F B = q ( v × B )
Drift velocity of the charge ⇒ v \Rightarrow v ⇒ v
Force on each charge = q v B =q v B = q v B
Charge density (charge number / unit volume) = n =n = n
Volume = A.l
→ \rightarrow → → \rightarrow → Force on a Current Carrying Conductor → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Charge Per Unit Length
Number of charge present in the unit of length l = n A l
Total Charge = n A l V
Total force on the length l
Current I = charge flowing per unit time = n A v q
Force and Torque due to Magnetic Field → \rightarrow → → \rightarrow → Charge Per Unit Length → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Force Due to Magnetic Field
Force on a Current Carrying Conductor → \rightarrow → → \rightarrow → Force Due to Magnetic Field → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Force Due to Magnetic Field
B ⃗ = B k ^ \vec{B}=B\hat{k} B = B k ^
v ⃗ = v sin ϕ ı ^ + v cos ϕ k ^ \vec{v}=v \sin \phi \hat{\imath}+v\operatorname{cos} \phi \hat{k} v = v sin ϕ ^ + v cos ϕ k ^
Force on one charge =q v ⃗ × B ⃗ q \vec{v} \times\vec{B} q v × B
= q ( v sin ϕ ı ^ + v cos ϕ k ^ ) × B k ^ q(v \sin \phi \hat{\imath}+v\operatorname{cos} \phi \hat{k})\times B\hat{k} q ( v sin ϕ ^ + v cos ϕ k ^ ) × B k ^
= q vsin ϕ ( ı ^ × k ^ ) \sin \phi(\hat{\imath}\times\hat{k}) sin ϕ ( ^ × k ^ )
= -qvB sin ϕ ȷ ^ \sin \phi\hat{\jmath} sin ϕ ^
Charge Per Unit Length → \rightarrow → → \rightarrow → Force Due to Magnetic Field → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Total Force
Total charge in a length l = nAlq
Total force on a length l = -nAlq v B sin ϕ ȷ ^ \phi\hat{\jmath} ϕ ^
Current I = nAvq
Total force on a length l = -IBl s i n ϕ j ^ sin \phi\hat{j} s in ϕ j ^
F ⃗ = I ( l ⃗ × B ⃗ ) \vec{F}=I(\vec{l} \times \vec{B}) F = I ( l × B )
l ⃗ = l sin ϕ i ^ + l cos ϕ k ^ \vec{l}=l \sin\phi \hat{i} + l \cos\phi \hat{k} l = l sin ϕ i ^ + l cos ϕ k ^
Force Due to Magnetic Field → \rightarrow → → \rightarrow → Total Force → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Force Between Two Current Carrying Conductor
Force on wire 2 due to wire 1
B 1 = μ 0 I 1 2 π d B_1=\frac{\mu_0 I_1}{2 \pi d} B 1 = 2 π d μ 0 I 1
F 21 = I 2 l μ 0 I 1 2 π d F_{21}=I_2 l \frac{\mu_0 I_1}{2 \pi d} F 21 = I 2 l 2 π d μ 0 I 1 = μ 0 I 1 I 2 2 π d l =\frac{\mu_0 I_1 I_2}{2 \pi d} l = 2 π d μ 0 I 1 I 2 l
Force Due to Magnetic Field → \rightarrow → → \rightarrow → Force Between Two Current Carrying Conductor → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Force On Wire
F 21 = μ 1 I 1 I 2 2 π d F_{21}=\frac{\mu_1 I_1 I_2}{2 \pi d} F 21 = 2 π d μ 1 I 1 I 2
B 2 = μ 0 I 2 2 π d B_2=\frac{\mu_0 I_2}{2 \pi d} B 2 = 2 π d μ 0 I 2
Force on wire 1
F 12 = I 1 l μ 0 I 2 2 π d = μ 0 I 1 I 2 2 π d l F_{12}=I_1 l \frac{\mu_0 I_2}{2 \pi d} = \frac{\mu_0 I_1 I_2}{2 \pi d} l F 12 = I 1 l 2 π d μ 0 I 2 = 2 π d μ 0 I 1 I 2 l
Force per unit length of wire 1 =μ 0 I 1 I 2 2 π d \frac{\mu_0 I_1 I_2}{2 \pi d} 2 π d μ 0 I 1 I 2
Total Force → \rightarrow → → \rightarrow → Force On Wire → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Example
Parallel current attract each other
Anti parallel currents repeal each other
I 1 = I 2 = 5 A I_1=I_2=5A I 1 = I 2 = 5 A
d = 10 c m = 10 − 2 m d=10\mathrm{cm} = 10^{-2}\mathrm{m} d = 10 cm = 1 0 − 2 m
F = μ 0 I 1 I 2 2 π d = 4 π × 10 − 7 × 5 × 5 2 π × 10 − 2 F = \frac{\mu_0 I_1 I_2}{2 \pi d} = \frac{4 \pi \times 10^{-7} \times 5 \times 5}{2 \pi \times 10^{-2}} F = 2 π d μ 0 I 1 I 2 = 2 π × 1 0 − 2 4 π × 1 0 − 7 × 5 × 5 = 5 × 10 − 4 N / m = 5 \times 10^{-4} N/m = 5 × 1 0 − 4 N / m
Force Between Two Current Carrying Conductor → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Torque on a Current Carrying Loop
Torque on a current carrying loop placed in a uniform magnetic field
Rectangle loop of sides a and b carrying a current I.
Loop is placed in a magnetic field B ⃗ \vec{B} B
Force On Wire → \rightarrow → → \rightarrow → Torque on a Current Carrying Loop → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Example
B ⃗ = B sin ϕ j ^ + B cos ϕ k ⃗ \vec{B}=B \sin \phi \hat{j} + B \cos \phi \vec{k} B = B sin ϕ j ^ + B cos ϕ k
Example → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Example
Force on 1
l ⃗ = b i ^ \vec{l} = b \hat{i} l = b i ^
F ⃗ 1 = I ( l ⃗ × B ⃗ ) = I b i ^ × ( B sin ϕ j ^ + B cos ϕ k ^ ) \vec{F}_1 = I(\vec{l} \times \vec{B}) = I b \hat{i} \times(B \sin \phi \hat{j} + B \cos \phi \hat{k}) F 1 = I ( l × B ) = I b i ^ × ( B sin ϕ j ^ + B cos ϕ k ^ )
= I b B sin ϕ k ^ − I b B cos ϕ j ^ = I b B \sin \phi \hat{k} - I b B \cos \phi \hat{j} = I b B sin ϕ k ^ − I b B cos ϕ j ^
Force on 2
l ⃗ = a j ^ = a j ^ \vec{l} = a \hat{j} = a \hat{j} l = a j ^ = a j ^
F ⃗ 2 = I ( l ⃗ × b ⃗ ) = I a j ^ × ( B sin ϕ j ^ + B cos ϕ k ^ ) \vec{F}_2 = I(\vec{l} \times \vec{b}) = I a \hat{j} \times (B \sin \phi \hat{j} + B \cos \phi \hat{k}) F 2 = I ( l × b ) = I a j ^ × ( B sin ϕ j ^ + B cos ϕ k ^ )
Torque on a Current Carrying Loop → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Example
Force on 3
l ⃗ = − b i ^ \vec{l} = -b \hat{i} l = − b i ^
F ⃗ 3 = I ( l ⃗ × B ⃗ ) = − I b i ^ × ( B sin ϕ j ^ + B cos ϕ k ^ ) \vec{F}_3 = I(\vec{l} \times \vec{B}) = -I b \hat{i} \times (B \sin \phi \hat{j} + B \cos \phi \hat{k}) F 3 = I ( l × B ) = − I b i ^ × ( B sin ϕ j ^ + B cos ϕ k ^ )
= − I b B sin ϕ k ^ + I b B cos ϕ j ^ = -I b B \sin \phi \hat{k} + I b B \cos \phi \hat{j} = − I b B sin ϕ k ^ + I b B cos ϕ j ^
Force on 4
l ⃗ = − a j ^ \vec{l} = -a \hat{j} l = − a j ^
F ⃗ 4 = − I a b cos ϕ i ^ \vec{F}_4 = -I a b \cos \phi \hat{i} F 4 = − I ab cos ϕ i ^
Example → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Torque due to Force
Total force = F 1 ⃗ + F ⃗ 2 + F ⃗ 3 + F ⃗ 4 = 0 =\vec{F_1} + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 = 0 = F 1 + F 2 + F 3 + F 4 = 0
Torque due to F ⃗ 1 \vec{F}_1 F 1
τ ⃗ 1 = r ⃗ 1 × F ⃗ 1 \vec{\tau}_1 = \vec{r}_1 \times \vec{F}_1 τ 1 = r 1 × F 1
= − a 2 j ^ × ( I b B sin ϕ k ^ − I b B cos ϕ j ^ ) = -\frac{a}{2} \hat{j} \times (I b B \sin \phi \hat{k} - I b B \cos \phi \hat{j}) = − 2 a j ^ × ( I b B sin ϕ k ^ − I b B cos ϕ j ^ )
= − I b B a sin ϕ i ^ = \frac {-I b B}{a} \sin \phi \hat{i} = a − I b B sin ϕ i ^
Example → \rightarrow → → \rightarrow → Torque due to Force → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Torque due to Force
Torque due to F 3 F_{3} F 3
τ 3 = r ⃗ 3 × F ⃗ 3 \tau_{3} =\vec{r}_3 \times \vec{F}_3 τ 3 = r 3 × F 3
= a 2 j ^ × ( − I b B sin ϕ k ^ + I b B cos ϕ j ^ ) = − I a b B 2 sin ϕ i ^ = \frac{a}{2} \hat{j} \times (-I b B \sin \phi \hat{k} + I b B \cos \phi \hat{j}) = -\frac{I a b B}{2} \sin \phi \hat{i} = 2 a j ^ × ( − I b B sin ϕ k ^ + I b B cos ϕ j ^ ) = − 2 I ab B sin ϕ i ^
Total torque τ ⃗ = τ 1 ⃗ + τ ⃗ 3 \vec{\tau} = \vec{\tau_1}+\vec{\tau}_3 τ = τ 1 + τ 3
= − I a b B sin ϕ i ^ = -I a b B \sin \phi \hat{i} = − I ab B sin ϕ i ^
τ ⃗ = I a b B sin ϕ ( k ^ × i ^ ) \vec{\tau} = I a b B \sin \phi (\hat{k} \times \hat{i}) τ = I ab B sin ϕ ( k ^ × i ^ )
Example → \rightarrow → → \rightarrow → Torque due to Force → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Magnetic Dipole Moment
Magnetic dipole
Dipole moment
A ⃗ = a b k ^ \vec{A}=a b \hat{k} A = ab k ^
Current = I
m ⃗ = I a b k ^ \vec{m}=Iab\hat{k} m = I ab k ^
B ⃗ = B sin ϕ j ^ + B cos ϕ k ^ \vec{B} = B \sin \phi \hat{j}+B \cos \phi \hat{k} B = B sin ϕ j ^ + B cos ϕ k ^
m ⃗ × B ⃗ = I a b k ^ × ( B sin ϕ j ^ + B cos ϕ k ^ ) \vec{m} \times \vec{B} =I a b \hat{k} \times(B \sin \phi \hat{j}+B \cos \phi \hat{k}) m × B = I ab k ^ × ( B sin ϕ j ^ + B cos ϕ k ^ )
= I a b B sin ϕ ( k ^ × j ^ ) =I a b B \sin \phi(\hat{k} \times \hat{j}) = I ab B sin ϕ ( k ^ × j ^ )
Torque due to Force → \rightarrow → → \rightarrow → Magnetic Dipole Moment → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Torque
τ ⃗ = m ⃗ × B ⃗ \vec{\tau}=\vec{m} \times \vec{B} τ = m × B
Torque on an electric dipole τ ⃗ = p ⃗ × E ⃗ \vec{\tau}=\vec{p} \times \vec{E} τ = p × E
Loop has N turns (closely arounds)
m ⃗ = N I A k ^ \vec{m}=N I A \hat{k} m = N I A k ^
Torque due to Force → \rightarrow → → \rightarrow → Torque → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Moving Coil Galvanometer
Magnetic Dipole Moment → \rightarrow → → \rightarrow → Moving Coil Galvanometer → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Moving Coil Galvanometer
Torque → \rightarrow → → \rightarrow → Moving Coil Galvanometer → \rightarrow → → \rightarrow →
Force And Torque Due To Magnetic Field L-1 Thank You
Moving Coil Galvanometer → \rightarrow → → \rightarrow → Thank You → \rightarrow → → \rightarrow →
Resume presentation
Force And Torque Due To Magnetic Field L-1 Force and Torque due to Magnetic Field $\rightarrow$ $\rightarrow$ Force and Torque due to Magnetic Field $\rightarrow$ Force on a Current Carrying Conductor $\rightarrow$ Charge Per Unit Length
