4 long wires each carrying current I as shown in the figure are placed at the points A , B , C and D . Find the magnitude and direction of: (i) Magnetic field at the centre of square (ii) Force per meter at point D .

(i) The magnitude of the magnetic field at the centre due to all wires will be same, i.e., B A = B B = B C = B D = B = μ 0 I 2 π 2 a And the direction of magnetic field is shown in the figure, The net magnetic field along the X -axis will be equal and opposite, so there will be no magnetic field along this axis. The magnetic field along the Y -axis is given by, B Net = B C + B A cos 45 ° + B D + B B cos 45 ° B Net = 4 B cos 45 ° = 4 × μ 0 I 2 π 2 a × 1 2 = μ 0 I π a (ii) Parallel current attract each other and antiparallel current repel each other, so The force per unit length will be given as, F A D = F D C = F 1 = μ 0 I 2 2 π 2 a and F B D = F 2 = μ 0 I 2 2 π 2 2 a Hence, the magnitude of the force along X -axis will be, F x = - F 1 + F 2 cos 45 ° = - μ 0 I 2 2 π 2 a + μ 0 I 2 2 π 2 2 a × 1 2 F x = - 3 μ 0 I 2 8 π a And along Y -axis, F y = - F 1 + F 2 sin 45 ° = - μ 0 I 2 2 π 2 a + μ 0 I 2 2 π 2 2 a × 1 2 = - μ 0 I 2 8 π a The magnitude of force will be, F = F x 2 + F y 2 = 10 μ 0 I 2 8 π a The force is given as, F → = - 3 μ 0 I 2 8 π a i ^ - - μ 0 I 2 8 π a j ^ .

An electron moving with velocity v → 1 = 1 i ^ m / s at a point in a magnetic field experiences a force F 1 → = - e j ^ N , where e is the charge of electron. If the electron is moving with a velocity v 2 → = i ^ - j ^ m / s at the same point, it experiences a force F 2 → = - e ( i ^ + j ^ ) N . . Find the magnetic field and the force the electron would experience if it were moving with a velocity v 3 → = v 1 → × v 2 → at the same point.

Case1: Velocity of electron v 1 ⇀ = 1 i ^ ms - 1 Force on electron F 1 → = - e j ^ N case 2: v 2 → = i ^ - j ^ ms - 1 F 2 → = - e ( i ^ + j ^ ) N Let us suppose the magnetic field B → = B x i ^ + B y j ^ + B z k ^ We know that, F → = q v → × B → . . . ( 1 ) F 1 → = q v 1 → × B → ⇒ - e j ^ = e 1 i ^ × ( B x i ^ + B y j ^ + B z k ^ ) ⇒ - e j ^ = e B x ( i ^ × i ^ ) + B y ( i ^ × j ^ ) + B z ( i ^ × k ^ ) ⇒ - j ^ = B y k ^ - B z j ^ ⇒ B y = 0 , B z = 1 Similarly, F → 2 = e ( v → 2 × B → ) ⇒ - e ( i ^ + j ^ ) = e ( ( i ^ - j ^ ) × ( B x i ^ + B y j ^ + B z k ^ ) ) ⇒ - ( i ^ + j ^ ) = B y k ^ - B z j ^ + B x k ^ - B z i ^ ⇒ i ^ + j ^ = B z i ^ + B z j ^ - ( B y + B x ) k ^ ⇒ B y + B x = 0 ∵ B y = 0 ⇒ B x = 0 Hence, the magnetic field B → = 1 k ^ T Now, calculate for v 3 → = v 1 → × v → 2 ⇒ v 3 → = ( i ^ × ( i ^ - j ^ ) ) ⇒ v 3 → = - k ^ ms - 1 On substitute the value of B → and v 3 → in equation 1 F 3 → = e v 3 → × B → ⇒ F 3 → = e ( - k ^ × k ^ ) ⇒ F 3 → = 0 N

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The figure shows a conductor of weight $1.0 N$ & length $l = 0.5 m$ placed on a rough inclined plane making an angle $30 °$ with the horizontal so that conductor is perpendicular to a uniform horizontal magnetic field of induction $B = 0.10 T$ . The coefficient of static friction between the conductor and the plane is $0.1$ . A current of $I = 10 A$ flows through the conductor inside the plane of this paper as shown. What is the force needed to be applied parallel to the inclined plane to keep the conductor at rest?

Important Questions on Magnetic Effect of Current

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Alpha Question Bank for Engineering: Physics>Chapter 25 - Magnetic Effect of Current>Exercise - 4>Q 4

$4$ long wires each carrying current $I$ as shown in the figure are placed at the points $0.88$ and $D$ . Find the magnitude and direction of:

(i) Magnetic field at the centre of square

(ii) Force per meter at point $D$ .

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Alpha Question Bank for Engineering: Physics>Chapter 25 - Magnetic Effect of Current>Exercise - 4>Q 5

The current density

\vec{J}

inside a long solid cylindrical wire of radius

a = 12 mm

is in the direction of the central axis and its magnitude varies linearly with radial distance

r

from the axis according to

J = \frac{J_{0} r}{a}

where,

J_{0} = \frac{10^{5}}{4 π} A m^{- 2}

. Find the magnitude of the magnetic field at

r = \frac{a}{2}

μT

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Alpha Question Bank for Engineering: Physics>Chapter 25 - Magnetic Effect of Current>Exercise - 4>Q 6

A neutral particle is at rest in a uniform magnetic field

B

. At

t = 0

, particle decays into two particles each of mass

m

and one of them having charge

q

. Both of these move off in separate paths lying in plane perpendicular to

\vec{B}

. At later time, the particles collide. Find this time of collision neglecting the interaction force.

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Alpha Question Bank for Engineering: Physics>Chapter 25 - Magnetic Effect of Current>Exercise - 4>Q 7

A non-uniform magnetic field $\vec{B} = B_{0} (1 + \frac{y}{d}) (- \hat{k})$ is present in a region of space in between $y = 0$ and $y = d$ . The lines are shown in the diagram. A particle of mass $m$ and positive charge $q$ is moving. Given an initial velocity $\vec{v} = v_{0} \hat{i}$ . Find the components of velocity of the particle when it leaves the field.

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Alpha Question Bank for Engineering: Physics>Chapter 25 - Magnetic Effect of Current>Exercise - 4>Q 8

A neutral atom of atomic mass number

100

which is stationary at the origin in gravity free space emits an

α

-particle (

A

) in

z

-direction. The production is

P

. A uniform magnetic field exists in the

x

-direction. Disregard the electromagnetic interaction between

A

and

P

. If the angle of rotation of

A

after which

A

and

P

will meet for the first time is

\frac{n π}{25}

radians, what is the value of

n

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Alpha Question Bank for Engineering: Physics>Chapter 25 - Magnetic Effect of Current>Exercise - 4>Q 9

In the figure shown a positively charged particle of charge $q$ and mass $m$ enters into a uniform magnetic field of strength $B$ as shown in the figure. The magnetic field points inwards and is present only within a region of width $d$ . The initial velocity of the particle is perpendicular to the magnetic field and $ϕ = 30 º$ . Find the time spent by the particle inside the magnetic field if $d = 0.2 m, B = 1 T, q = 1 C$ , $m = 1 kg$ and $v = 1 m s^{- 1}$ . Use $\sin 45 ° = 0.7$ , if required.

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Alpha Question Bank for Engineering: Physics>Chapter 25 - Magnetic Effect of Current>Exercise - 4>Q 10

A thin beam of charged particles is incident normally on the boundary of a region containing a uniform magnetic field as shown. The beam comprise of mono energetic $α$ and $β^{-}$ particles, each possessing a kinetic energy $T$ . Neglecting interaction between particles of the beam, find the separation between the points on line $PQ$ where the $α$ and $β$ particles emerge out of the magnetic field. (express your answer in terms of $T, B, α (2 e, m_{α}) & β^{-} (- e, m_{e}),$ where all terms have their usual meanings.)

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Alpha Question Bank for Engineering: Physics>Chapter 25 - Magnetic Effect of Current>Exercise - 4>Q 11

An electron moving with velocity

{\vec{v}}_{1} = 1 \hat{i} m / s

at a point in a magnetic field experiences a force

\vec{F_{1}} = - e \hat{j} N,

where e is the charge of electron. If the electron is moving with a velocity

\vec{v_{2}} = \hat{i} - \hat{j} m / s

at the same point, it experiences a force

\vec{F_{2}} = - e (\hat{i} + \hat{j}) N .

. Find the magnetic field and the force the electron would experience if it were moving with a velocity

\vec{v_{3}} = \vec{v_{1}} \times \vec{v_{2}}

at the same point.

Important Questions on Magnetic Effect of Current

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