(1) Two straight long vertical wires W_{1} and W_{2} carry steady currents *I* and 3*I* respectively as shown. P is a point midway between the wires. The resultant magnetic flux density at P due to the currents in the wires is **B**. The direction of **B** is

(a) vertically upwards

(b) vertically downwards

(c) horizontal and directed normally into the plane of the figure (away from the reader)

(d) horizontal and directed normal to the plane of the figure, towards the reader

(e) horizontal and directed towards the wire W_{2}

The magnetic field at P due to the current *I* in the wire W_{1} is directed normally into the plane of the figure (away from the reader). The magnetic field at P due to the current 3*I* in the wire W_{2} opposes the above field. Since the magnitude of the field due to the wire W_{2} is three times that due to the wire W_{1}, the resultant field will be directed normal to the plane of the figure, towards the reader [Option (d)].

(2) In the above question if the current 3*I* in the wire W_{2} is switched off, what will be the magnetic field at P?

(a) **B**

(b) – **B**

(c) – **B**/2

(d) **B**/2

(e) – **B**/3

The direction of the magnetic field produced by the current *I* in the wire W_{1} is *opposite *to the net field **B **in question No.(1). So the sign has to be *negative*. Since the currents *I* and 3*I* in the wires W_{1} and W_{2} produce opposing fields at P (in question no.1), the resultant field **B** has magnitude proportional to 2*I*. When the current 3*I* in the wire W_{2} is switched off, the field at P is proportional to *I* so that the magnitude is reduced to half the net value in question No.(1).

The field at P in this case is therefore – **B**/2.

The following questions are specifically meant for ** AP Physics C** aspirants:

(3) The magnetic flux density at the centre of a plane circular coil of radius *R* carrying a current *I* is **B**_{0}. If the current in the coil is reversed, the magnetic flux density *on the axis of the coil* at distance *R* from the centre will be

(a) **B**_{0}/√8

(b) – **B**_{0}/√2

(c) – **B**_{0}/√8

(d) – **B**_{0}/2

(e) **B**_{0}/2

The magnitude of the magnetic flux density *B* at a point *on the axis of the coil* at distance *x* from the centre of the coil is given by

*B* = μ_{0}*nR*^{2}*I/*2(*R*^{2 }+ *x*^{2})^{3/2} where *n* is the number of turns in the coil.

The magnetic field *B*_{0 }at the centre (*x* = 0) of the coil is given by

*B*_{0} = μ_{0}*nI/*2*R*

The field (*B*_{1}) on the axis at distance *R* from the centre (*x* = R) is given by

*B*_{1} = μ_{0}*nR*^{2}*I/*2(*R*^{2 }+ *R*^{2})^{3/2} = μ_{0}*nR*^{2}*I/*4√2 *R*^{3}

Or, *B*_{1} = μ_{0}*nI/*4√2 *R*

Therefore, *B*_{1} = *B*_{0}/2√2 = *B*_{0}/√8

Since the current in the coil is reversed, the direction of the magnetic field must be reversed so that the correct option is – **B**_{0}/√8.

(4) An electron moving with velocity (30** i** +** **40** j**) ms^{–1} enters a magnetic field of – 2 **k** tesla where **i**,** j** and** k **are unit vectors along the x, y and z directions respectively. Then

(a) the speed of the electron will not change; but the path will become parabolic

(b) the speed alone will change

(c) the speed will not change; but the path will become helical

(d) the speed will change and the path will become circular

(e) the speed will not change; but the path will become circular

Outside the magnetic field the path of the electron is straight and is contained in the xy plane. The magnetic field is directed along the negative z-direction an hence the electron enters *normally* into the magnetic field. A magnetic force acts perpendicular to the direction of motion of the electron.

The path of the electron within the magnetic field will therefore be circular. But the *speed* will be unchanged since a stationary magnetic field cannot change the kinetic energy of the electron. The correct option is (e).

[If the initial velocity of the electron had an z-component also (for instance, **v** = 30** i** +** **40** j + 20 k**), the electron would enter the magnetic field at an angle other than 90º and then the path would be helical].

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